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pentagon [3]
3 years ago
9

How are force and motion related

Physics
1 answer:
sp2606 [1]3 years ago
5 0
Force is used to put things in motion you can’t have motion without force!
Even if it just pushing a piece of paper you still use force to put the paper in motion!
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Vedmedyk [2.9K]
Time stops everything is made out of atoms so if atoms freeze everything freezes
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3 years ago
What type of waves are sound waves, and how do they<br> transfer energy?
klemol [59]

Answer:

Sound wave types - longitudinal waves

Longitudinal waves - Vibrating string the creates sound in the way it moves.

Explanation:

Longitudinal waves have particles of the medium that are displaced in a parallel direction to energy transport.

7 0
3 years ago
What is the average velocity of the person taking this walk?
Butoxors [25]
5 would be the answer
6 0
3 years ago
Read 2 more answers
A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will
77julia77 [94]

Answer:

V=200cm^3\\\\\rho =0.500g/cm^3

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3

Moreover, since the density is computed via the division of the mass by the volume:

\rho =\frac{m}{V}

We obtain:

\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0
3 years ago
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load m_{LS} = 10000 kg

mass of flat bed m_{FB} = 20000 kg

initial speed of truck v_{0} = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs F_{N}     -------------let this be equation 1

where F_{N} = normal force = mg

so

Fs,max = μs mg

ma_{max} = μs mg

divide through by mass

a_{max} = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a_{max} = 0.5 × 9.8 m/s²

a_{max} = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

v_{f}² = v_{0}² + 2aΔx

where v_{f} is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0
3 years ago
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