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lawyer [7]
3 years ago
9

Describe an instance when the motion energy of an object changed, causing some other energy change as well.

Physics
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

For instance, an immobile object located at some height above the ground has potential energy. If the object starts to fall, its potential energy is converted to motion or kinetic energy. The heavier and the higher the object is, the more potential energy it will have. Transfer of thermal energy via conduction is aided by vigorously vibrating particles in a medium that has gained energy through heating.

Explanation:

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Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play .  But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t  + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

7 0
3 years ago
The question states: two large, parallel conducting plates are 12cm
AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, E=\frac{F}{q}

E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0
3 years ago
When an object is fully magnetized, all of its magnetic domains will be ?
kkurt [141]
The answer is to this question D
4 0
4 years ago
Read 2 more answers
A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct
labwork [276]

Answer:

Speed of the helium after collision = 246 m/s

Explanation:

Given that

Mass of helium ,m₁ = 4 u

u₁=598 m/s

Mass of oxygen ,m₂ = 32 u

u₂  = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

Speed of the helium after collision = 246 m/s

6 0
3 years ago
At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

5 0
4 years ago
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