Given that,
Length of bar = 600 mm
Diameter of bar = 40 mm
Diameter of hole = 30 mm
Length of hole = 100 mm
Modulus of elasticity = 85 GN/m²
Load = 180 kN
We need to calculate the area of cross section without hole
Using formula of area
[tex]A=\dfrac{\pi\times d^2}{4}[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi\times40^2}{4}[/tex]
[tex]A=1256.6\ mm^2[/tex]
We need to calculate the area of cross section with hole
Using formula of area
[tex]A=\pi\times\dfrac{(d_{b}^2-d_{h}^{2})}{4}[/tex]
Put the value into the formula
[tex]A=\pi\times\dfrac{(40^2-30^2)}{4}[/tex]
[tex]A=549.77\ mm^2[/tex]
We need to calculate the total contraction on the bar
Using formula of total contraction
Total contraction = contraction in bar without hole part + contraction in bar with hole part
[tex]Total\ contraction = \dfrac{F\times L_{1}}{A_{1}\times E}+\dfrac{F\times L_{2}}{A_{2}\times E}[/tex]
Where, F = load
L = length
A = area of cross section
E = modulus of elasticity
Put the value into the formula
[tex]Total\ contraction=\dfrac{180\times10^3}{85\times10^{3}}(\dfrac{500}{1256.6}+\dfrac{100}{549.77})[/tex]
[tex]Total\ contraction = 1.227\ mm^2[/tex]
Hence, The total contraction on the bar is 1.227 mm²
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.0 m/s
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball
Answer:
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Applying the impulse-momentum equation;
Impulse = change in momentum
Ft = m∆v
F = (m∆v)/t
Where;
F = force
t = time
m = mass
∆v = v2 - v1 = change in velocity
Given;
m = 0.80 kg
t = 0.050 s
The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.
v2 = 25 m/s
v1 = -25 m/s
∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s
Substituting the values;
F = (m∆v)/t
F = (0.80×50)/0.05
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
A cylindrical capacitor is made of two concentric cylinders. The inner cylinder has radius r1 = 4 mm, and the outer one a radius r2= 8 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference V = 4 V is applied between the inner and outer cylinder?
Answer:
E = 9.62*10^-8 J
Explanation:
The energy stored in a capacitor is given by the following formula:
[tex]E=\frac{1}{2}CV^2[/tex] (1)
E: energy stored
C: capacitance
V: potential difference of the capacitor = 4 V
The capacitance for a concentric cylindrical capacitor is:
[tex]C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}[/tex] (2)
L: length of the capacitor = 150m
r2: radius of the outer cylinder = 8mm = 8*10^-3m
r1: radius of the inner cylinder = 4mm = 4*10^-3m
εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2
You replace the expression (2) into the equation (1) and replace the values of all parameters:
[tex]E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J[/tex]
The energy stored in the cylindrical capacitor is 9.62*10-8 J
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1150 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of ?9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
(a) For what time interval is the rocket in motion above the ground?
(b) What is its maximum altitude?
(c) What is its velocity just before it hits the ground?
Before the engines fail, the rocket's altitude at time t is given by
[tex]y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
and its velocity is
[tex]v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
The rocket then reaches an altitude of 1150 m at time t such that
[tex]1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
Solve for t to find this time to be
[tex]t=11.2\,\mathrm s[/tex]
At this time, the rocket attains a velocity of
[tex]v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}[/tex]
When it's in freefall, the rocket's altitude is given by
[tex]y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity, and its velocity is
[tex]v_2(t)=124\dfrac{\rm m}{\rm s}-gt[/tex]
(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for [tex]y_2(t)[/tex] to reach 0:
[tex]1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s[/tex]
So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.
(b) Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]
where [tex]v_f[/tex] and [tex]v_i[/tex] denote final and initial velocities, respecitively, [tex]a[/tex] denotes acceleration, and [tex]\Delta y[/tex] the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means [tex]y_2[/tex] will contain the information we need to find the maximum height.
[tex]-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)[/tex]
Solve for [tex]y_{\rm max}[/tex] and we find that the rocket reaches a maximum altitude of about 1930 m.
(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to [tex]y_2(t)[/tex]) to be about 32.6 s. Plug this into [tex]v_2(t)[/tex] to find the velocity before it crashes:
[tex]v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}[/tex]
That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.
A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from left to right, through a uniform magnetic field, with the plane of the loop remaining perpendicular to the plane of the paper at all times. Determine the direction of the current induced in the loop as it swings past the locations labeled (a) I and (b) II. Specify the direction of the current in terms of the points x, y, and z on the loop (e.g., x→y→z or z→y→x). The points x, y, and z lie behind the plane of the paper. What is the direction of the induced current at the locations (c) I and (d) II when the loop swings back, from right to left?
Complete Question
The complete question iws shown on the first uploaded image
Answer:
a
[tex]y \to z \to x[/tex]
b
[tex]x \to z \to y[/tex]
Explanation:
Now looking at the diagram let take that the magnetic field is moving in the x-axis
Now the magnetic force is mathematically represented as
[tex]F = I L[/tex] x B
Note (The x is showing cross product )
Note the force(y-axis) is perpendicular to the field direction (x-axis)
Now when the loop is swinging forward
The motion of the loop is from y to z to to x to y
Now since the force is perpendicular to the motion(velocity) of the loop
Hence the force would be from z to y and back to z
and from lenze law the induce current opposes the force so the direction will be from y to z to x
Now when the loop is swinging backward
The motion of the induced current will now be x to z to y
Two wires carry current I1 = 47 A and I2 = 29 A in the opposite directions parallel to the x-axis at y1 = 9 cm and y2 = 15 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.
Explanation:
Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:
The region under the first wire, where [tex]y < y_1[/tex];The region between the first and the second wire, where [tex]y_1 < y < y_2[/tex]; andThe region above the second wire, where [tex]y > y_2[/tex].Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.
In the region below the first wire, [tex]B_1[/tex] points out of the [tex]x y[/tex]-plane while [tex]B_2[/tex] points into the [tex]x y[/tex]-plane.In the region between the two wires, both [tex]B_1[/tex] and [tex]B_2[/tex] point into the [tex]x y[/tex]-plane.In the region above the second wire, [tex]B_1[/tex] points into the [tex]x y[/tex]-plane while [tex]B_2[/tex] points out of the [tex]x y[/tex]-plane.The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.
At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:
[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],
where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.
Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.
In the region under the first wire, [tex]R_1 = y_1 - y = 9 - y[/tex], while [tex]R_2 = y_2 - y = 15 - y[/tex]. In the region above the second wire, [tex]R_1 = y - 9[/tex], while [tex]R_2 = y - 15[/tex].Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.
Strength of the magnetic field due to the first wire: [tex]\displaystyle B_1 = \frac{\mu\, I_1}{4\pi\, R_1}[/tex].Strength of the magnetic field due to the second wire: [tex]\displaystyle B_1 = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:
[tex]B_1 = B_2[/tex].
[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].
Simplify this equation:
[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].
In the region under the first wire (where [tex]y < 9[/tex],) this equation becomes [tex]\displaystyle \frac{47}{9 - y} = \frac{29}{15 - y}[/tex].In the region above the second wire (where [tex]y > 15[/tex],) this equation becomes [tex]\displaystyle \frac{47}{y - 9} = \frac{29}{y - 15}[/tex].These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.
A lens of focal length 12cm forms an
three times the size of the
to the object. The distance between the object and the image is what
b) 16 cm
Magnification, m = v/u
3 = v/u
⇒ v = 3u
Lens formula : 1/v – 1/u = 1/f
1/3u = 1/u = 1/12
-2/3u = 1/12
⇒ u = -8 cm
V = 3 × (-8) = -24
Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm
Calculate the Reynold's number using a viscosity of air as 1.81E-05 kilograms/(meters-seconds), the density of air (see above), the diameter as 0.15 m, and, from the data, 0.89 m/s.
Answer:
8924.6
Explanation:
We are given that
Viscosity of air,[tex]\eta=1.81\times 10^{-5}kg/m-s[/tex]
Density of air,[tex]\rho=1.21kg/m^3[/tex]
Diameter,d=0.15 m
v=0.89m/s
We have to find the Reynold's number.
Reynold's number,R=[tex]\frac{\rho vd}{\eta}[/tex]
Substitute the values then we get
[tex]R=\frac{1.21\times 0.89\times 0.15}{1.81\times 10^{-5}}[/tex]
R=[tex]8924.6[/tex]
Hence, the value of Reynold's number=8924.6
Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed vx must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Express your answer in meters per second to two significant figures.
Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
Answer:
The answer is 3.0
Explanation:
If a 2,000-kg car hits a tree with 500 N of force over a time of 0.5 seconds,
what is the magnitude of its impulse?
O A. 14,700 kg-m/s
B. 2,500 kg-m/s
C. 250 kg-m/s
D. 10,000 kg-m/s
Answer:
c 250 kg-m/s
Explanation:
happy to help!!
Answer: 250 kg
Explanation:
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800 J/m3, what is the value of the dielectric constant k?
Answer: The value of the dielectric constant k = 1.8
Explanation:
If C= ε A/d and
Electrostatic energy W = 1/2CV^2
Substitutes C in the first formula into the energy formula.
W = 1/2 ε A/d × V^2
Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed
Substitute V = Ed into the energy W.
W = 1/2 × ε A/d ×( Ed )^2
W = 1/2 × ε A/d × E^2 × d^2
d will cancel one of the ds
W = 1/2 × ε Ad × E^2
W/Ad = 1/2 × ε × E^2
W/V = 1/2 × ε E^2
Where Ad = volume V
E = dielectric strength
εo = permittivity of free space = 8.84 x 10^-12 F/m
W/V = 2800 J/m^3
Let first calculate the dielectric strength
2800 = 1/2 × 8.84×10^-12 × E^2
5600 = 8.84×10^-12E^2
E^2 = 5600/8.84×10^-12
E = sqrt( 6.3 × 10^14)
E = 25 × 10^7
75% of E = 18.9 × 10^6Jm
The permittivity of the material will be achieved by using the same formula
2800 = 1/2 × ε E^2
2800 = 0.5 × ε × (18.9×10^6)^2
2800 = ε × 1.78 × 10^14
ε = 2800/1.78×10^14
ε = 1.57 × 10^-11
Dielectric constant k = relative permittivity
Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is
k = 1.57×10^-11/8.84×10^-12
k = 1.776
k = 1.8 approximately
Therefore, the value of the dielectric constant k is 1.8
Make predictions for the following five situations based on what you observed in part A. Check your answers only after you have made all five predictions. 1. The magnet is turned so that the south pole is near the wire while the switch is closed. Prediction Observation: 1. 2. The leads to the power supply are reversed (consider both orientations of the magnet). Prediction: Observation: 3. The north pole of the magnet is held near the wire but the switch remains open. Prediction: Observation 4. The north pole of the magnet is held: (a) closer to the wire and (b) farther from the wire. Prediction: Observation 5. The magnet is turned so that it is parallel to the wire while the switch is closed. Prediction Observation Resolve any discrepancies between your predictions and your observations. (Hint: Recall the right- hand rule used to determine the direction of the force exerted by a magnetic field on moving charges.) II. The magnetie field of a current-carrying wire A. Suppose you place a small magnet in a magnetic field and allow it to rotate freely. How will the magnet orient relative to the external magnetic field lines?
Answer:
the answer is D.
Explanation: Mark me brainlest
Calculate the average speed of a car between 0 And 10 second give the unit
Answer:
I think the distance should be given in the question because the formula for getting the
average speed= total distance/total time taken
The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polarized in the vertical direction is incident from the left and has an average intensity of 32 W/m2. Determine the average intensity of the light that emerges on the right in the drawing (a) when sheet A alone is removed, (b) when sheet B alone is removed, (c) when sheet C alone is removed, and (d) when sheet D alone is removed.
Answer:
Explanation:
When sheet A is removed
[tex]I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2[/tex]
When sheet B is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2[/tex]
When sheet C is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2[/tex]
When sheet D is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_B=32\cos^230=24W/m^2\\\\I_C=24\cos^260=6W/m^2[/tex]
The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate
(i) the instantaneous velocity when t = 1
(ii) the instantaneous acceleration when t =1
Answer:
(i) v = 44 m/s
(ii) a = 72 m/s^2
Explanation:
You have the following equation for the potion of a car:
[tex]x=20t+6t^4[/tex]
(i) The instantaneous velocity is the derivative of x in time:
[tex]\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3[/tex]
for t = 1 is:
[tex]v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s[/tex]
(ii) The instantaneous acceleration is the derivative of the velocity:
[tex]\frac{dv}{dt}=24(3)t^2=72t^2[/tex]
for t = 1
[tex]a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2[/tex]
There are seven ___ included in the periodic table.
we have seven groups in the periodic table
The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat
capacities (C ) of water and ice are 75.3 and 37.7JK'mol', respectively. Calculate
AH for the freezing of 1 mol of supercooled water at -10.0°C.
Answer:
[tex]\Delta H_{tot} = 2258.025\,kJ[/tex]
Explanation:
The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of [tex]25^{\circ}C[/tex]. Then:
[tex]\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}[/tex]
[tex]\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})[/tex]
Finally, the amount of heat released from water is now computed by replacing variables:
[tex]\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right][/tex][tex]\Delta H_{tot} = 2258.025\,kJ[/tex]
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
On putting estimate values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]
⇒ [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]
⇒ [tex]v=1.94 \ cm[/tex]
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]
⇒ [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]
⇒ [tex]{h_{1}}=1.16 \ mm[/tex]
A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.
Required:
a. Determine the direction of the force on a charge.
b. Determine the direction of the force on a charge.
1. along the line between the charge and the center of the square outward of the center
2. along the line between the charge and the center of the square toward the center
3. along the side of the square toward the other charge that lies on the side
4. along the side of the square outward of the other charge that lies on the side
Answer:
Explanation:
Given that:
length of side , a = 0.5 m
charge , q = 6.65 mC
length of diagonal , d = 0.5 * sqrt(2)
d = 0.707 m
F is the force due to adjacent particle ,
F1 is the force due to diagonal particle
Now , for the net charge on a particle
Fnet = 2 * F * cos(45) + F1
Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2
Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)
Fnet = 3.05 *10^6 N
the magnitude of net force acting on each particle is 3.05 *10^6 N
part B)
for the direction of particle
d) along the line between the charge and the center of the square outward of the center
A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm2 and a large one of cross-sectional area 300 cm2. What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg
Answer:
735.75 N
Explanation:
We are given;
Area of small lift; A1 = 5 cm²
Area of large lift; A2 = 300 cm²
Mass required to lift large lift;m = 4500 kg
Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N
By Pascal's law,
F1/A1 = F2/A2
We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.
Thus;
F1 = (F2 × A1)/A2
Plugging in the relevant values, we have;
F1 = (44145 × 5)/300
F1 = 735.75 N
A 1,150-kg car moving east at 80 km/h collides head-on with a 1,900-kg car moving west at 40 km/h , and the two cars stick together. It’s initial speed is 5km/h. How much KE is lost in the collision?
Answer:
Ask a pro,bro
Explanation:
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an object with momentum must also have
a. acceleration
b. kinetic energy
c. impulse
d. force
An object with momentum must also have impulse.
An object with momentum must also have: c. impulse.
Momentum can be defined as the product of the mass possessed by an object and its velocity. Also, it is a vector quantity and as such has both magnitude and direction.
Mathematically, momentum is giving by the formula;
[tex]Momentum = Mass \times Velocity[/tex]
In Physics, the impulse experienced by an object is always equal to the change in momentum of the object. This is due to the force acting on an object.
[tex]Impulse = Change\;in\;momentum\\\\Force \times time = m \Delta V[/tex]
In conclusion, an object with momentum must also have impulse.
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A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 5.90 m: (a) the initially stationary spelunker is accelerated to a speed of 2.30 m/s; (b) he is then lifted at the constant speed of 2.30 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 69.0 kg rescue by the force lifting him during each stage
Answer:
W₃ = 3310.49 J , W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
A sample of a compound weighs 2.54 grams. how many ounces (oz.) is this equivalent to?
a. 71.9 oz
b. 0.0139 oz
c. 0.0896 oz
d. 30.8 oz
The correct answer is C. 0.0896 oz
Explanation:
Both grams and ounces are units use to measure the mass of objects. Additionally, 1 gram represents a smaller amount of mass than the one represented by 1 ounce. Indeed, 1 ounce is equivalent to 28.349 grams and therefore, 1 gram is equivalent to 0.035 ounces ( 1 ounces x 1 gram / 28.349 grams = 0.03527 ounces).
According to this, you can convert grams to ounces by multiplying the amount of grams given by 0.035, which is the equivalent of 1 gram to ounces. This means, to determine how many ounces there are in 2.54 you multiply it by 0.035 which is equivalent to 0.0896 ounces (2.54 x 0.03527 = 0.0896 oz).
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Good luck on your assignment
A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s after this, the ball reaches the bottom of the incline. (a) What was the magnitude of the ball's acceleration, assume it to be constant
Answer:
a) a = 2.35 m/s^2
Explanation:
(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
x: position of the ball after t seconds = 87 m
t: time = 8.6 s
a: acceleration of the ball = ?
vo: initial velocity of the ball = 0 m/s
You solve the equation (1) for a:
[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]
You replace the values of the parameters in the previous equation:
[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]
The acceleration of the ball is 2.35 m/s^2
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
V₀ₓ = 10.94 m/s
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
V₀y = 18.87 m/s
In which of the following is negative work performed by an individual?
A) A mover picks up a packing box up off the floor.
B) A person leans against a wall.
C) A shopper lowers a can of beans from a store shelf into the grocery cart.
Answer:
c
Explanation:
Calcular el módulo del vector resultante de dos vectores fuerza de 9 [N] y 12 [N] concurrentes en un punto o, cuyas direcciones forman un ángulo de a) 30˚ b) 45˚ y c) 90˚
Answer:
a) 20.29N
b) 19.43N
c) 15N
Explanation:
To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.
In all these calculations you can asume that one of the vectors coincides with the x-axis.
a)
[tex]F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N[/tex]
b)
[tex]F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N[/tex]
c)
[tex]F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N[/tex]
The motion of a free falling body is an example of __________ motion
Answer:
accelerated
Explanation:
The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.
Newton's second law of motion
