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Vesna [10]
3 years ago
5

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

tion on the crate compared to your push
Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

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A current of 6.93 A 6.93 A in a long, straight wire produces a magnetic field of 4.15 μT 4.15 μT at a certain distance from the
natka813 [3]

Answer:

0.334 m

Explanation:

The magnetic field due to current in a straight thin wire, B, is given by

B = \dfrac{\mu_0 I}{2\pi R}

where \mu_0 is the permeability of free spave with value 4\pi\times10^{-7} and R is the distance from the wire.

R = \dfrac{\mu_0 I}{2\pi B}

Substituting the values from the question,

R = \dfrac{4\pi\times10^{-7}\times 6.93}{2\pi\times 4.15\times 10^{-6}}

R = \dfrac{2\times10^{-1}\times6.93}{4.15} = 0.334 \text{ m}

7 0
4 years ago
Read 2 more answers
A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to
Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\\or\\P=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_{1}*v_{1})-F*t=(m_{1}*v_{2})

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]

3 0
3 years ago
A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta
Blababa [14]

Explanation:

Given that,

The slope of the ramp, \theta=8^{\circ}

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

W=\dfrac{1}{2}m(v^2-u^2)=0

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

W=Fd\ cos\theta

Here, \theta=90+8=98^{\circ}

W=mgd\ cos\theta

W=60\times 9.8\times 300\ cos(98)

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

W_f=-W

W_f=24550.13\ J

Hence, this is the required solution.

6 0
3 years ago
You push against a steamer trunk with a force of 750 N at an angle of 25° with the horizontal . The trunk is on a flat floor and
djverab [1.8K]
This means that the horizontal force is 750sin(25°). To be able to move the truck, force applied must be greater than static friction, which equals to its coefficient (0.77) x normal contact force (= weight)

Hence, 750sin(25°) = 0.77mg. m = 750sin(25°)/(0.77g)
3 0
3 years ago
I can fly but have no wings. I can cry but I have no eyes. Wherever I go, darkness follows me. What am I?
nlexa [21]

cloud?? thats prob wrong lol

7 0
3 years ago
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