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Vesna [10]
3 years ago
13

An elephant travels 16metes to the left in 20s, what is the average velocity?​

Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

S=16metes

t=20sec

V=?

V=S/t

V=16/20

V=0.8m/s

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What is the mass of a stone moving at a speed of 15 meters/second and having a momentum of 7.5 kilogram - meters / seconds
Elanso [62]

Answer:

0.5 kg

Explanation:

The momentum of an object is defined as

p = mv

where

m is the mass

v is the velocity

In this problem we have,

v = 15 m/s is the velocity of the stone

p = 7.5 kg m/s is the momentum

Solving for m, we can find the mass of the stone:

m=\frac{p}{v}=\frac{7.5 kg m/s}{15 m/s}=0.5 kg

4 0
3 years ago
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If two items are equal in size, the denser one will
Talja [164]
Be heavier

density=mass÷volume

if two items have the same size they have the same volume so the heavier one will be the denser one
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3 years ago
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
Which of the following is not true?
ahrayia [7]
The answer is either C or D.. 

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Where is the sun's energy most concentrated
Maru [420]

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator (0^{\circ}), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

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