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riadik2000 [5.3K]
2 years ago
14

A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,

and the box remains a rest. The box rests on a horizontal surface that has some friction with the box. The normal force on the box due to the floor is equal to
Physics
1 answer:
Kipish [7]2 years ago
3 0

Answer:

its on wheels and they are supposed to make it eas

Explanation:

You might be interested in
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0
olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0
3 years ago
Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?
Leto [7]
The answer is 19.3 j
3 0
2 years ago
Read 2 more answers
Dominic made the table below to organize his notes about mixtures.
Andrei [34K]

Answer:

Althought it us possible to have more then one state, it's also possible to have only one state of matter

Explanation:

You can make solutions of only one state if matter, for example , it two liquids can be mixed to form a solution they are called miscible.

4 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh
MrMuchimi

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

W=\vec F \cdot \vec r

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

W=F.d

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

W=350\ N\ .\ 2\ m=700\ Joule

4 0
3 years ago
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