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A 4.0 kg shot put is thrown with 30 N of force. What is its acceleration?

weeeeeb [17]

7.5 m/s

a = F ÷ m

a = (30 N) ÷ (4.0 kg)

a = 7.5 m/s2

7 0

3 years ago

You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?

Snezhnost [94]

If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.

7 0

3 years ago

Two identical bodies are sliding toward each other on a frictionless surface. One moves at 1 m/s and the other at 2 m/s. They co

mario62 [17]

Answer:

$V=\dfrac{3}{2}\ m/s$

Explanation:

Two identical bodies are sliding toward each other on a frictionless surface.

Initial speed of body 1, m₁ = 1 m/s

Initial speed of body 2, m₂ = 2 m/s

They collide and stick.

We need to find the speed of the combined mass. Let V is the speed of the combined mass.

Using the conservation of momentum.

$m_1u_1+m_2u_2=V(m_1+m_2)$

We have, m₁ = m₂ = m

$m\times 1+m\times 2=(m+m)V\\\\m+2m=2m\times V\\\\3m=2mV\\\\V=\dfrac{3}{2}\ m/s$

So, the speed of the combined mass is $\dfrac{3}{2}\ m/s$ .

6 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

What happens to parallel light rays that strike a concave lens?

torisob [31]

Answer:

They diverge on refraction

Explanation:

When parallel light rays strike a concave lens, they will diverge that is they spread out .

Concave lens is also known as diverging lens, which means that when parallel rays of light strike on it, the lens spreads out the light rays ( that is it diverges the rays of light) that are refracted through it.

At the middle of concave lens is thinner.

When light is passes through the lens they diverge it or spread out.

The concave lens causes light rays to bend away or diverge from its axis since the concave lens is a diverging lens.

5 0

3 years ago