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Goshia [24]
3 years ago
13

The answer please it’s very simple this is 7th grade science

Physics
2 answers:
olganol [36]3 years ago
7 0

Answer:

Kinetic Energy

Explanation:

The explanation is the cart is moving and the people are not so the answer is kinetic energy.

allochka39001 [22]3 years ago
3 0

The greatest point for kinetic is at the bottom and in the middle it is in half and at the top it is at the highest in potential energy.

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A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc
Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g=\frac{15}{1000}=0.015 kg

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

\frac{1}{2}(m+M)^2V^2=(m+M)gh

Using g=9.8m/s^2

Substitute the values

\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086

V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}

V=\sqrt{2\times 3.015\times 9.8\times 0.086}}

V=1.3m/s

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

mv+Mv'=(m+M)V

Using the formula

0.015v+3(0)=3.015(1.3)

0.015v=3.015(1.3)

v^2=\frac{3.015(1.3)}{0.015}=261.3

v=261.3 m/s

5 0
4 years ago
A ball is thrown off the top of a building and lands on the ground below.
natita [175]

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

K=\frac{1}{2} mv^2

where m is the mass of the object, and v is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

  • mass of the ball
  • velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball  before touching the gound can be determined.

4 0
3 years ago
A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
3 0
3 years ago
Hii I need help ASAP with this physics question
Readme [11.4K]

Between the top of the first and the top of the second loop, the coaster has lost potential energy = mgh, where h = 22.2 - 15 = 7.2m

This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.

4 0
3 years ago
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl
bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

230=0+\dfrac{1}{2}\times9.8\times t^2

t=\sqrt{\dfrac{230\times2}{9.8}}

t=6.85 sec

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0
3 years ago
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