All categories

3 years ago

What is the SPEED of a ball that travels 120 m in 6s?

Physics

1 answer:

AVprozaik [17]3 years ago

7 0

Answer:

20m/second

Explanation:

The reason the answer is 20m/second is because to find the speed of the ball in this question you have to divide the distance over the time giving you the result of 20m/second

You might be interested in

You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

5 0

3 years ago

Element A and element B have bonded and formed compound AB. Which of the following statements is true? Compound AB has chemical

creativ13 [48]

The correct answer from the choices listed above is the first option. The statement that is true would be that c<span>ompound AB has chemical and physical properties that are completely different from those of A and B. They completely different substances with different properties.</span>

6 0

3 years ago

Read 2 more answers

PLEASE Please HELP ME...

cupoosta [38]

Answer:

girl this easy ask yo teacher for help lol

7 0

2 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

2 years ago

A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.

Klio2033 [76]

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

#SPJ1

3 0

1 year ago