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Nonamiya [84]
3 years ago
13

A horizontal force of 90.0 N is required to push a 75kg object along a horizontal surface at a constant speed. What is the magni

tude of the force of friction. Please show your work I am having troubles.
Physics
1 answer:
Varvara68 [4.7K]3 years ago
6 0
If the object is moving at a constant speed, acceleration is 0. So:

F = ma

F = m\times 0

F = 0

The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so 

Friction = 90 N
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3 years ago
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If the resistance of a circuit increases, that means that the current has to
Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

  • Current is <em>inversely proportional</em> to the resistance.
  • from the relation given below, we can clearly see the relation between current and resistance;

                              V=IR

                              I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0
3 years ago
An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is
Fynjy0 [20]

Answer: 2.55meter

Explanation: Using the second equation of motion.

S{hieght} = U*t + {g*t²}/2

Where U is initial velocity =0m/s

g is acceleration due to gravity 10m/s²

t is time 1secs

So we have,

hieght = 0 + {g*t²}/2

hieght = {10*(1)²}/2

Total hieght travelled is 10/2

Which is 5 meter.

But we are asked to find the hieght above the window which as a hieght of 2.45meter.

So,

hieght above window would be

{5 - 2.45}meter

Which is 2.55 meter.

8 0
3 years ago
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an
disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
3 years ago
In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the f
zhannawk [14.2K]

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

8 0
3 years ago
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