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2 years ago

Which force below acts at 90 degrees and is applied by a surface/wall to an object.

Physics

1 answer:

Molodets [167]2 years ago

4 0

Answer:

B. normal

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An ideal spring is hung vertically from the ceiling. The spring constant is k = 125 N/m. A block of mass m = 650 g (1000 g = 1 k

Brrunno [24]

Answer:

0.102 m

Explanation:

k = spring constant of the spring = 125 N/m

m = mass of the block attached to the spring = 650 g = 0.650 kg

x = maximum extension of the spring

h = height dropped by the block = x

Using conservation of energy

Spring potential energy gained = Gravitational potential energy lost

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) (125) x = (0.650) (9.8)

x = 0.102 m

3 0

3 years ago

Alkali metals _____.

Lyrx [107]

I think the answer is C

7 0

3 years ago

2.What is the weight of an object that has a mass of 5 kg?<br>

Artemon [7]

Answer:

5 kg

or 11 lbs

Explanation:

4 0

3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

4 years ago

Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the

inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change

4 0

3 years ago