Answer:
The correct answer will be "2 m".
Explanation:
As we know,
⇒ [tex]\frac{Mirror's \ size}{Body's \ height} =\frac{1}{2}[/tex]
Now,
⇒ [tex]Size \ of \ a \ mirror = \frac{n}{2}[/tex]
Then,
⇒ [tex]Size \ of \ mirror = \frac{4}{2}[/tex]
⇒ [tex]=2 \ meter[/tex]
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:
I just ryhmed
There are seven ___ included in the periodic table.
we have seven groups in the periodic table
How does the engine get the spacecraft to space?
Answer:
An electric power source is used to ionize fuel into plasma. Electric fields heat and accelerate the plasma while the magnetic fields direct the plasma in the proper direction as it is ejected from the engine, creating thrust for the spacecraft.
Explanation:
A charge q1 of -5.00 × 10‐⁹ C and a charge q2 of -2.00 × 10‐⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a third charge of +15.0 × 10‐⁹ C.
P.S. 10-⁹ is 10^-9
Answer:
Explanation:
The equilibrium position will be in between q₁ and q₂ .
Let this position of third charge be x distance from q₁
q₁ will pull the third charge towards it with force F₁
F₁ =9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²
= 675 x 10⁻⁹ / x²
q₂ will pull the third charge towards it with force F₂
F₂ =9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ /( .40-x )²
= 270 x 10⁻⁹ / ( .40-x )²
For equilibrium
675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / ( .40-x )²
5 / x² = 2 / ( .40-x )²
( .40-x )² / x² = 2/5 = .4
.4 - x / x = .632
.4 - x = .632x
.4 = 1.632 x
x = .245 .
24.5 cm
so third charge must be placed at 24.5 cm away from q₁ charge.
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
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
A coin is placed 17.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 26.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?
Answer: The coefficient of static friction between the coin and the turntable is 0.13.
Explanation:
As we know that,
Centripetal force = static frictional force
[tex]\frac{mv^{2}}{r} = F_{s}[/tex]
or, [tex]\frac{mv^{2}}{r} = \mu_{s} \times m \times g[/tex]
v = [tex]\sqrt{\mu_{s} \times r \times g}[/tex]
or, [tex]\mu_{s} = \frac{v^{2}}{rg}[/tex] ......... (1)
Here, it is given that
r = 17 cm, [tex]\omega[/tex] = 26 rpm,
and v = [tex]r \omega[/tex] ..........(2)
Putting equation (2) in equation (1) we get the following.
[tex]\mu_{s} = \frac{r^{2}\omega^{2}}{rg}[/tex]
= [tex]\frac{17 \times 10^{-2} \times (26 \times [\frac{2 \times \pi}{60}]^{2})}{9.8}[/tex]
= 0.128
= 0.13 (approx)
Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.
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
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:
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
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.
Read more: https://brainly.com/question/15517471
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
What is the meaning of powee
Hope it helps
Good luck on your assignment
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.
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]
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
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
Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a and y=(2)/(5)h
Explanation:
The x and y coordinates of the center of mass are:
xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA
ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA
Assuming uniform density, the center of mass is also the center of area.
xcm = ∫ x dA / ∫ dA = ∫ x y dx / A
ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A
First, let's find the area:
A = ∫ y dx
A = ∫₀ᵃ (-h/a² x² + h) dx
A = -⅓ h/a² x³ + hx |₀ᵃ
A = -⅓ h/a² (a)³ + h(a)
A = ⅔ ha
Now, let's find the x coordinate of the center of mass:
xcm = ∫ x y dx / A
xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)
xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)
xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)
xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)
xcm = (¼ ha²) / (⅔ ha)
xcm = ⅜ a
Next, we find the y coordinate of the center of mass:
ycm = ∫ y² dx / A
ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)
ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)
ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)
ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)
ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)
ycm = ⅖ h
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:
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 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 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.
6. A capacitor of charge 3 x 10 coulomb has a potential of 50volts. What is the capacitance of the capacitor?
Answer:
Explanation:
Sry
Answer:
C = Q/V
where C is capacitance, Q is charge and V is voltage
C = (3×10)/50
C = 30/50
= 0.6F where F is in Farads
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
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
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
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]
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
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]
