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3 years ago

Running for his life, Riley travels 100m in 9.83s. What was his average velocity in

1 answer:

5 0

100m / 9.83 sec = 10.17 m/s (rounded)

There are 3,600 seconds in an hour.

(100 m / 9.83 sec) x ( 3600 sec/hr) = 36,622.6 m/hr = 36.62 km/hr

That's about 22.8 miles per hour.

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What is the acceleration of the box?

worty [1.4K]

Answer:

a=4,32m/s^2

Explanation:

Fnet = F1 - F2

= 12-1.2

= 10.8N

m=2.5kg

Fnet =ma

10.8=2.5a then divide both sides by 2.5 to get acceleration

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3 years ago

A scientist records the motion of an object. She writes down that the

Phantasy [73]

The answer is Velocity

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3 years ago

1. What happens to particle spacing when temperature increase, What we call this process?

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Answer:

Particle spacing increases and it's called evaporating

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3 years ago

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago

Charges of +2 µC and +3 µC are 4 mm from each other. Raina’s group draws four diagrams trying to represent the electrical force

ahrayia [7]

The correct diagram is shown below:

The charges of +2 µC and +3 µC are 4 mm from each other. The diagram below represents the electrical force between the charges. i.e. repulsive force. However the force of repulsion exerted by charge +3 µC on +2 µC will be more. The same charges repel each other and opposite charges attract each other.

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3 years ago

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