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3 years ago

A 6.2 kg block is pushed across a horizontal surface

Physics

1 answer:

sergey [27]3 years ago

5 0

Answer:

6 m/s²

Explanation:

We are given;

The mass of the block as 6.2 kg
Horizontal force as 50.9 N
Frictional force as 13.7 N

We are required to determine the acceleration of the block.

According to second Newton's Law of motion; Resultant force is directly proportional to the rate of change in linear momentum.

Therefore;

F = Ma , where F is the resultant force, m is the mass and a is the acceleration.

In this case;

Resultant force, F = Horizontal force - friction force

= 50.9 N - 13.7 N

= 37.2 N

Thus;

37.2 N = 6.2 kg × a

Hence;

acceleration, a = 37.2 N ÷ 6.2 kg

= 6 m/s²

Therefore, the acceleration of the block is 6 m/s²

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a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball

MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .

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3 years ago

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When sedimentary rock is exposed to heat and pressure, what does it change into?

patriot [66]

Metamorphic rock this possess often occurs in the mantle

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3 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

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2 years ago

Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled?

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Ans) A) Centripetal force will be doubled.

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That means centripetal force is directly proportional to the mass of the particle

So, if we double the mass, centripetal force will be increased by twofolds.

So, option A) is correct.

Now, looking at the other options,

B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.

Option C) and D) reduces centripetal force which are also not possible here.

So, only Option A) is correct

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3 years ago

35 grados fahrenheit es mayor a 99 grados celsius

lord [1]

Answer:

Ok so holi pues Mira

Explanation:

35 C = 95 F

asi q falso :)

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