Ask question

Ask question

All categories

2 years ago

5

Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less

er height than the previous bounce before coming to rest in place.
In one or two sentences, explain how the experiment shows the direction that gravity pulls an object, specifically showing that it does not pull an object parallel to the ground.

1 answer:

Vladimir [108]2 years ago

6 0

Answer:

What do you mean?

Explanation:

You might be interested in

A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat

mihalych1998 [28]

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

6 0

3 years ago

Help me with Economics please and thank you the question is going to be down in a attached file while the answers on here

Tresset [83]

I think is A or B it depends on like what the trying to answer

6 0

3 years ago

Read 2 more answers

What condition is necessary for an object to make a good reference point?

777dan777 [17]

The object is fixed relative to the motion you are trying to describe.

8 0

4 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago