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polet [3.4K]
3 years ago
12

A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

ion is 0.25. What is the force of friction on the box? What is the acceleration of the box?
Physics
2 answers:
fgiga [73]3 years ago
5 0

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

Burka [1]3 years ago
4 0
Yeah what that guy said lol
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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in
Kazeer [188]

Answer:

Explanation:

it equals16 2211

8 0
3 years ago
Acceleration is zero if
tester [92]
It’s doesn’t change meaning it’s 0
4 0
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Can someone tell me the answer and the reason
spin [16.1K]

Answer:

(a) the blocks all had different masses.

Explanation:

The surface is smooth, therefore coefficient of friction is tending to zero.

Forces for each blocks varied from 6N to 8N to 7N to 5N

The blocks were made of different materials and different materials are going to have varying weight for the same size of block.

8 0
3 years ago
3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)
lesantik [10]

Answer:

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

Explanation:

Given:

  • length of the concrete highway, l=12\ m
  • coefficient of thermal expansion, \alpha=12\times 10^{-6}\ ^{\circ}C^{-1}
  • range of temperature variation, (-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C

<u>Now form the equation of thermal expansion:</u>

\Delta l=l\times \alpha\times \Delta T

\Delta l=12\times (12\times 10^{-6})\times 80

\Delta l=1.152\times 10^{-2}\ m

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

\Delta x=2\times \Delta l

\Delta x=2\times( 1.152\times 10^{-2})

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

4 0
3 years ago
Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal
NeX [460]

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

8 0
3 years ago
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