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3 years ago

The force that opposes the start of motion is referred to as _____.

1 answer:

7 0

The force that opposes the start of motion is referred to as static friction.

Friction is a force that opposes motion in general, and static friction opposes initial movement. This is because, when two objects are stopped near one another, weak bonds begin to form between the atoms on their surfaces. These bonds must be broken in order for one object to move away from the other. This resistance caused by the bonds between motionless objects is static friction.

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As shown in the diagram below, a 1 kg rock tied to a rope is

Tatiana [17]

Answer:

Explanation:

Angular momentum has a formula of L = mvr. Fillingin:

L = (1.0)(5.0)(1.0)

L = 5.0 kg*m/s

4 0

3 years ago

Can i get help for the parts for the two questions? MATHPHYSSSS

Anestetic [448]

Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))

F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

P = 0.218 N / 0.712 m²

P = 0.306 N/m²

7 0

3 years ago

Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.

GalinKa [24]

It IS PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J

6 0

3 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

Can anyone help??????????????

DerKrebs [107]

I think the answer should be the last one. Magnets attract magnets with unlike poles and repel magnets with like poles

3 0

4 years ago