Ask question

Ask question

All categories

3 years ago

9

A crow drops .11 -kg clam onto a rocky beach from a height of 9.8 m. What is the kinetic energy of the clam when it is 5.0 m abo

ve the ground? What is it's speed at that point?

1 answer:

Alchen [17]3 years ago

5 0

Answer:

Kinetic Energy=5.17J

Speed=9.7m/sec

Explanation:

Given the mass of the clam is m=0.11kg

The height from which the rock was thrown is h=9.8m

Given the rock is at 5m above the ground.

The distance travelled by the rock is (s)=9.8-5=4.8m

The initial velocity is u=0m/sec

let the final velocity be v m/sec

We know that the acceleration of the particle is $a=9.8m/sec^{2}$

We know that $v^{2} -u^{2} =2as$

$v^{2} -0^{2} =2\times 9.8\times 4.8$

$v^{2} =94.08$

v=9.7m/sec

Now the kinetic energy= $\frac{1}{2}mv^{2}$

KE= $\frac{1}{2}\times 0.11\times 9.7^{2}$ =5.17J

You might be interested in

Define newtons first law of motion with examples(class 9 ) plssss itz urgent

Akimi4 [234]

<h3>Newtons first law state that if a body is in motion it will be in motion and if a body is in rest it will tend to be in rest. This phenomena is also called INERTIA. Example: We tend to fall sideways when a car turn suddenly</h3>

5 0

3 years ago

Read 2 more answers

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?

Monica [59]

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

$\boxed{\sf{\bold{W = F \times s}}}$

With the following condition :

W = work (J)
F = force (N)
s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

$\boxed{\sf{\bold{W = F \times \Delta h}}}$

With the following condition :

W = work (J)
F = force (N)
$\sf{\Delta h}$ = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

$\sf{W = F \times \Delta h}$

$\boxed{\sf{\bold{W = m \times g \times \Delta h}}}$

With the following condition :

W = work (J)
m = mass of the object (kg)
g = acceleration of the gravity (m/s²)
$\sf{\Delta h}$ = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

F = force = 100 N
$\sf{\Delta h}$ = change of altitude 300 m

What was asked :

W = work = ... J

Step by step :

$\sf{W = F \times \Delta h}$

$\sf{W = 100 \times 300}$

$\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}$

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

<h3>See More :</h3>

Work that he had done to lift object brainly.com/question/26341717
Converting work to potential energy brainly.com/question/26487284

5 0

2 years ago

A coaxial cable is 10 meters long, and is filled with lossless Teflon, having a relative permittivity of 2.1. There is a matched

sattari [20]

My milkshake brings all the boys to the yard

And they're like, it's better than yours

Dam right it's better than yours

I can teach you, but I have to charge

My milkshake brings all the boys to the yard

And they're like, it's better than yours

Dam right it's better than yours

I can teach you, but I have to charge

I know you want it

The thing that makes me

What the guys go crazy for

They lose their minds

The way I wind

I think it's time

La la, la la, la

Warm it up

La la, la la, la

The boys are waiting

La la, la la, la

Warm it up

La la, la la, la

The boys are waiting

My milkshake brings all the boys to the yard

And they're like, it's better than yours

Dam right it's better than yours

I can teach you, but I have to charge

My milkshake brings all the boys to the yard…

6 0

3 years ago

Relative density has no unit .why ?

Klio2033 [76]

Answer:

Relative density is a ratio of the density of a certain material to the density of a reference material. Relative density has no unit, because the units of the material in question cancel out with the units of the reference material. For example, say a cooking oil’s density is 0.9 g/cm^3.

Explanation:

7 0

3 years ago