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blondinia [14]
2 years ago
11

A elephant kicks a 5.0 kg stone with 150 J of kinetic energy.

Physics
1 answer:
romanna [79]2 years ago
6 0
(1/2)(5.0)(v²) = 150
v= 7.75m/s


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SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
HELP PLEASE
4vir4ik [10]

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
7 0
3 years ago
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Which of the following are results of the force of gravity?
docker41 [41]
Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.
7 0
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Please need help on this
Nookie1986 [14]
The answer is c) 80°c
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2 years ago
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A diverging lens of focal length 18.0m is used to view a shark that is 90.0m away from the lens. If the image formed is 1.0m lon
Reil [10]

Answer:

i. + 22.5 m ii. 4.0 m

Explanation:

i. Image distance

Using the lens formula

1/u + 1/v = 1/f where f = focal length = + 18.0 m, u = object distance = distance of shark away from lens = + 90.0 m and v = image distance from lens = unknown

So, we find v

1/v = 1/f - 1/u

= 1/+18 - 1/+90

= (5 - 1)/90

= 4/90

v = 90/4

= + 22.5 m

So the image is real and formed 22.5 m away on the other side of the lens.

ii Length of Shark

Using the magnification formula, m = image height/object height = image distance/object distance. image height = 1.0 m where object height = length of shark.

m = image distance/object distance

= v/u

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0.25 = image height/object height

So,

object height = image height/0.25

= 1.0 m/0.25

= 4.0 m

So, the length of the shark is 4.0 m

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