The magnitude of the acceleration of the crate is 1.033 m/s².
The given parameters;
weight of the crate, W = 925 Nhorizontal force applied to the carte, F = 325 Nangle of inclination of the force, θ = 25⁰the coefficient of friction, μ = 0.25The mass of the crate is calculated as;
W = mg
[tex]m = \frac{W}{g} \\\\m = \frac{925}{9.8} \\\\m = 94.388 \ kg[/tex]
The normal force on the crate is calculated as;
Fₙ = 925 - 325 x sin(25)
Fₙ = 925 - 137.35
Fₙ = 787.65 N
The frictional force on the object is calculated as follows;
[tex]F_k = \mu F_n\\\\F_k = 0.25 \times 787.65\\\\F_k = 196.91 \ N[/tex]
The magnitude of the crates acceleration is calculated from the net horizontal force on the crate;
[tex]\Sigma F_x = 0\\\\Fcos (\theta) - F_k = ma\\\\325\times cos(25) \ - \ 196.91 = ma\\\\97.54 = ma[/tex]
[tex]a = \frac{97.54}{94.388} \\\\a = 1.033 \ m/s^2[/tex]
Thus, the magnitude of the acceleration of the crate is 1.033 m/s².
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help pls i don’t know what to dooooooo
Explanation:
distance from ground
mass
amount of compression
Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surface. Determine the total mechanical energy of the river water per unit mass (in kJ/kg) and the power generation potential of the entire river at that location (in MW). The density of water is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2. The total mechanical energy of the river per unit mass is kJ/kg. The power generation potential of the entire river at that location is MW..
Answer:
1. 0.574 kJ/kg
2. 315.7 MW
Explanation:
1. The mechanical energy per unit mass of the river is given by:
[tex] E_{m} = E_{k} + E_{p} [/tex]
[tex] E_{m} = \frac{1}{2}v^{2} + gh [/tex]
Where:
Ek is the kinetic energy
Ep is the potential energy
v is the speed of the river = 3 m/s
g is the gravity = 9.81 m/s²
h is the height = 58 m
[tex] E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg [/tex]
Hence, the total mechanical energy of the river is 0.574 kJ/kg.
2. The power generation potential on the river is:
[tex] P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW [/tex]
Therefore, the power generation potential of the entire river is 315.7 MW.
I hope it helps you!
Supply the missing force necessary to achieve equilibrium. Show your work.
Analysing the Question:
We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it
In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N
Finding the missing force:
Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces
So, (106 + 106 + 142 + x) = 496
354 + x = 496
x = 496 - 354
x = 142 N
Therefore, the missing force is 142 N
Average velocity of Mike Phelps swimming 100 m race in the 50 m long pool (2 laps) is approximately equal to *
A. 0 m/s
B. 1 m/s
C. 2 m/s
D. 4 m/s
Answer:
2
Explanation:
A mass (m = 30 g) falls onto a spring (k = 7.3 N/m) from a height (h = 25 cm). The spring compresses an additional amount x before temporarily coming to a stop. What is the value of x?
Answer:
x₁ = 0.1878 m
Explanation:
For this exercise we will use conservation of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point with fully compressed spring
Em_f = K_e + U
Em_f = ½ K x² + m g x
energy is conserved
Em₀ = Em_f
m g h = ½ K x² + m g x
½ K x² + mg (x- h) = 0
let's substitute
½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0
3.65 x² + 0.294 (x- 0.25) = 0
x² + 0.080548 (x- 0.25) = 0
x² - 0.020137 + 0.080548 x = 0
x² + 0.080548 x - 0.020137 = 0
let's solve the quadratic equation
x = [0.080548 ±√ (0.080548² + 4 0.020137)] / 2
x = [0.080548 ± 0.29502] / 2
x₁ = 0.1878 m
x₂ = -0.1072 m
These are the compression and extension displacement of the spring
Matter is anything that takes up space and has mass. O A. True O B. False
Answer:True
Explanation:Matter is everything around you. Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).
41
Adam is pushing his box of baseball
equipment with a force of 10 N and
the box is pushing back towards
Adam with a force of 6 N. What is the
total net force? What will happen to the motion of
the box? Explain.
The Magnolia loh
deneaker notes
Answer:
16
Explanation:
6+10=16
the box will go forward but it will be a little harder.
Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
Answer:
significant
Explanation:
The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).
A particle moves along a path described by y=Ax^3 and x = Bt, where tt is time. What are the units of A and B?
Answer:
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
Explanation:
From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:
[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)
[tex][l] = B\cdot [t][/tex] (Eq. 2)
Now we finally clear each constant:
[tex]A = \frac{[l]}{[l]^{3}}[/tex]
[tex]A = \frac{1}{[l]^{2}}[/tex]
[tex]B = \frac{[l]}{[t]}[/tex]
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
The magnetic field at the center of a 0.60-cm-diameter loop is 2.7 mT . What is the current in the loop
Answer:
12.898A
Explanation:
The formula for calculating the magnetic field in the loop bus expressed as;
B = Iμ0/2r
Given
Diameter d = 0.60cm
Radius r = d/2 = 0.30cm
r = 0.0030m
Permittivity of free space μ0 = 4π×10^-7
Magnetic field strength B = 2.7×10^-3T
Substitute into the formula and get I
2.7×10^-3 = 4π×10^-7I/2(0.0030)
2.7×10^-3 = 4π×10^-7I/0.0060
Cross multiply
2.7×10^-3 × 6.0×10^-3 = 4π×10^-7I
16.2×10^-6 = 4π×10^-7I
I = 16.2×10^-6/4π×10^-7
I = 16.2×10^-6/4(3.14)×10^-7
I = 16.2×10^-6/12.56×10^-7
I = 1.2898×10^{-6+7}
I = 1.2898×10
I = 12.898A
Hence the current in the loop is 12.898A
Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be?
Answer:
new height of ramp must be 4 times that of the first ramp.
Explanation:
From conservation of energy, we know that;
Potential energy at the top of ramp = kinetic energy at the bottom of ramp.
Thus;
mgh_t = ½mv²
m will cancel out to give;
gh_t = ½v²
Thus means that the height of the first ramp is directly proportional to the square of the speed.
Thus;
h_t ∝ v²
Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.
Thus;
h'_t ∝ (2v)²
h'_t ∝ 4v²
From earlier we saw that;
h_t ∝ v²
Thus;
New height of ramp is;
h'_t ∝ 4h_t
Thus, new height of ramp must be 4 times that of the first ramp.
To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
The given parameters;
velocity at the bottom ramp, = vlet the height of the ramp = hApply the principle of conservation of energy;
P.E = K.E
[tex]mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}[/tex]
To increase the speed 2v at the bottom of the ramp, the height will be;
[tex]H = \frac{(2v)^2}{2g} \\\\H = \frac{4v^2}{2g} \\\\H = 4(\frac{v^2}{2g} )\\\\H = 4(h)[/tex]
Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
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is this correct?? or wrong?
I am a cell. I am long and thin. I reach all the way from the brain
to the tip of a finger. I have a special coat of fat that helps me do
my job. My job is to send electrical signals from one part of the
body to another.
Answer:
Neurons
Explanation:
We humans have a nervous system that coordinates our behavior and transmits signals between different parts of our body.
Now, this nervous system contains a lot of nerve cells which we call Neurons. These Neurons have a cell like body and their job is to transmit signals from one part of our body to another.
Thus, the cell is called Neurons.
what is the meaning of relative as a noun?
Answer:
noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
Ос
А group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
KINEMATICS: MOTION ALONG STRAIGHT LINE
The nearest grocery is 60m, east from your house. You are walking at 1.2 m's for 15.Os toward
the grocery when it started to rain and you ran back to your house to get an umbrella. It took
you another 5.0s to go back to your house. You start walking again at 1.2m's until you reach
the grocery. (a) What is your average speed? (b) What is your average velocity?
Explanation:
hope this helps, cheers!
The values for the average speed and average velocity are;
(a) [tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]
(b) [tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]
The reason the above values are correct is given as follows;
The given parameters are;
Distance of the nearest grocery store to the house, d = 60 m
Direction of the grocery store = East from the house
Walking speed, s = 1.2 m/s
Time of walking before the rain started, t₁ = 15.0 s
Time it takes to go back to the house, t₂ = 5.0 s
Speed at which the start walking again, s = 1.2 m/s
(a) The formula for average speed, [tex]s_{avg}[/tex], is given as follows;
[tex]s_{avg} = \dfrac{Total \ distance }{Sum \ of \ time \ taken}[/tex]
The distance walked before the rain, d₁ = 1.2 m/s × 15.0 s = 18 meters
Distance covered on returning to the house = d₂ = d₁ = 18 meters
The total distance covered, ∑Distance = d₁ + d₂ + 60
∴ ∑Distance = 18 m + 18 m + 60 m = 96 m
The time, t₃, it takes to get to the grocery store on the second attempt after getting the umbrella is given as follows;
[tex]t_3 = \dfrac{d}{v}[/tex]
[tex]t_3 = \dfrac{60 \ m}{1.2 \ m/s} = 50.0 \, s[/tex]
The total time, ∑Time = t₁ + t₂ + t₃
∴ ∑Time = 15.0 s + 5.0 s + 50.0 s = 70.0 s
[tex]Average \ speed = \dfrac{\sum Distance}{\sum Time}[/tex]
[tex]Average \ speed = \dfrac{96 \, m}{70 \, s} = \dfrac{48}{35} \ m/s= 1.37\overline{142857} \ m/s[/tex]
[tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]
(b) The average velocity, [tex]v_{avg}[/tex], is given as follows;
[tex]v_{avg}= \dfrac{Total \ displacement}{Total \ time}[/tex]
The total displacement = The total change in location = The change in location from the house to the grocery = 60 m
The total time is the same as for the average speed, ∑Time = 70 s
[tex]\therefore v_{avg}= \dfrac{60 \, m}{70 \, s} = \dfrac{6}{7} \ m/s = 0.\overline{857142} \ m/s[/tex]
[tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]
Learn more about average speed and average velocity here:
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Write a conclusion to this activity in which you completely and intelligently describe the characteristics of an object that is traveling in uniform circular motion. Give attention to the quantities speed, velocity, acceleration and net force.
Answer:
This question is incomplete
Explanation:
This question is incomplete but there are some characteristics that are peculiar to an object traveling in uniform circular motion.
An object moving in a uniform circular motion generally has a constant speed with which it uses to move in a circle (the object moves round/tangent to the circle). This then means the object will move in different direction although with the same speed. This change in direction means the object will accelerate (inwards) at different velocities (since velocity is a vector quantity that measures both magnitude and direction). Because the object moves with different velocities; this makes the object an accelerating object.
From the descriptions above, it can be conceived/visualized that the net force acting on an object in a uniform circular motion is a centripetal force. This is because the net force acting on the object is directed towards the center of the circle the object in rotating/moving in. Without this net force, the object would have moved in a straight line and thus not changing it's direction. The formula used to calculate this centripetal force is
Fc = mv²/r
where Fc is the centripetal force
m is the mass of the object
v is the velocity of the object
r is the radius of the curvature/curved path
The net force on an object moving in a circular path is directed inwards and it is known as centripetal force. The centripetal force increases with increase in speed and acceleration of the object.
The acceleration of an object travelling in a circular path is directed inwards and the magnitude of the acceleration is given as;
[tex]a_c = \frac{v^2}{r} = \omega^2 r[/tex]
where;
v is the linear speed of the objectω is the angular speed of the objectr is the radius of the circle[tex]a_c[/tex] is the centripetal acceleration of the objectThe net force acting on the object is given as follows;
[tex]F_c = ma_c[/tex]
The net force on the object is directed inwards and it is known as centripetal force.
Learn more about circular motion here: https://brainly.com/question/14672628
If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the circuit?
Answer:
I = 0.96 A
Explanation:
No of electrons, [tex]n=1.8\times 10^{16}[/tex]
Time, t = 3 ms = [tex]3\times 10^{-3}\ s[/tex]
We need to find the electric current. We know that electric charge per unit time is equal to the electric current.
[tex]I=\dfrac{q}{t}[/tex]
q = ne (Quantization of electric charge)
[tex]I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A[/tex]
So, the electric current is 0.96 A.
A teacher pushes on a file cabinet sitting on a
level floor. Which force must the teacher
overcome if they want to slide the cabinet across
the floor: Gravity, Normal, or Friction?
Marking brainliest
A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?
Answer:
Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.
Explanation:
To find the distance at which the first package will land we need to calculate the time:
[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
Y(f) is the final position = 0
Y(0) is the initial position = 160 m
V(0y) is initial speed in "y" direction = 0
g is the gravity = 9.81 m/s²
t is the time=?
[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]
Now we can find the distance of the first package:
[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]
Then, after 2 seconds the distance traveled by plane is (from the initial position):
[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]
Now, the distance of the second package is:
[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]
The distance between the packages is:
[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]
Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.
I hope it helps you!
I NEED HELP PLEASEE ITS AN ECONOMICS QUESTION ABOVE
Answer:
I believe the answer is Property taxes
Explanation:
Answer: I'm pretty sure property taxes
Explanation:
Regardless of what state a substance is in, it is always that substance
True
False
Which state of matter is most similar to solids
Answer:
liquids
Explanation
A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.
Answer:
The wavelength is [tex]\lambda = 1.2 * 10^8 nm[/tex]
Explanation:
From the question we are told that
The frequency of operation of the microwave is [tex]f = 2.50 GHz = 2.50 *10^{9} \ Hz[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}[/tex]
=> [tex]\lambda = 0.12 \ m [/tex]
converting to nanometer
[tex]\lambda = 1.2 * 10^8 nm[/tex]
If we throw a body upwards at time t=0 with an initial speed s=25 m/s, what would the body's speed, position and velocity?
Answer:
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
Explanation:
This is a vertical throwing exercise.
The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.
Let's write the kinematics equations
v = v₀ - g t
v² = v₀² - g (y-y₀)
y = y₀ + v₀ t - ½ g t²
in our case the initial velocity is
v₀ = 25 m / s
and we zero the system at the launch point
y₀ = 0
the equations remain
v = 25 - 9.8 t
v² = 25² - 9.8 y
y = 0 + 25 t - ½ 9.8 t²
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what projection angle would produce the greatest downrange distance?
Answer:
45°
Explanation:
Range is the horizontal distance travelled by an object undergoing projectile motion.
Range is given by the formula:
[tex]R=\frac{u^2sin(2\theta)}{g}[/tex]
where u = velocity, g = acceleration due to gravity and θ = angle above the horizontal.
For angle of 15°:
[tex]R=\frac{u^2sin(2*15)}{g}=\frac{0.5u^2}{g}[/tex]
We get a maximum range when sin(2θ) = 1
sin(2θ) = 1
2θ = sin⁻¹(1)
2θ = 90⁰
θ = 90°/2
θ = 45⁰
For angle of 45°:
[tex]R=\frac{u^2sin(2*45)}{g}=\frac{u^2}{g}[/tex]
Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?
Answer:
12.5 m
Explanation:
The first thing we would do is to calculate the wavelength. To do this, we use the formula
v = fλ, where
v = wave speed
f = frequency
λ = wavelength
If we make wavelength the formula, we have
wavelength = speed / frequency
Now, we substitute the values we had been given and we have
wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m
half of this said wavelength will be
= 3.50 / 2
= 1.75 m
As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,
(10 / 1.75) = 5.7
So the separation will have to be 7 half wavelengths
= (7 * 1.75) = 12.5 m
A Navy Seal of mass 80 kg parachuted directly down into an enemy harbor. At one point while she was falling, the resistive force that air exerted on her was 520 N upward. What can you determine about her motion at this point in time
Answer:
The Navy Seal is accelerating downwards at the rate of 3.3 m/s²
Explanation:
Given;
mass of the Navy Seal, m = 80 kg
the upward resistive force on her, F = 520 N
Her net downward force is given by;
[tex]F_{net} = F_{down} - F_{up}\\\\F_{net} = (80*9.8) - 520\\\\F_{net} = 264 \ N[/tex]
Her downward acceleration at this time is given by;
F = ma
a = F / m
a = 264 / 80
a = 3.3 m/s²
Therefore, the Navy Seal is accelerating downwards at the rate of 3.3 m/s²
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
