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arlik [135]
2 years ago
15

Which forces are contact forces give one example of each​

Physics
1 answer:
insens350 [35]2 years ago
8 0

Answer:

A contact force is any force that requires contact to occur. Contact forces are ubiquitous and are responsible for most visible interactions between macroscopic collections of matter. Pushing a car up a hill or kicking a ball across a room are some of the everyday examples where contact forces are at work.

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What does a lunar eclipse and a solar eclipse have in common
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During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".
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3 years ago
It is 5.5 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h
stealth61 [152]

Explanation:

Displacement = 5 km

A.

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

= 25/9 m/s

Remember,

700 watt = 700 J/s

Velocity = displacement/time

Time = 5000/(25/9)

= 1800 s

Energy = power * time

= 700 * 1800

= 1,260,000

= 1260 kJ

B.

Converting km/h to m/s,

3 km/h * 1000 m/1 km * 1 h/3600 s

= 5/6 m/s

290 watt = 290 J/s

Velocity = displacement/time

Time = 5000/(5/6)

= 6000 s

Energy = power * time

= 290 * 6000

= 1,740,000

= 1740 kJ

C.

Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.

6 0
3 years ago
A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a tem
olya-2409 [2.1K]

Answer:

800j/k

Explanation:

6 0
3 years ago
Read 2 more answers
Which type of wave is more dangerous for humans? Microwaves or x-rays?
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X- rays. It can damage living tissue
3 0
3 years ago
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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
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