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Oxana [17]
2 years ago
10

Definition: The energy transferred by a force to a moving object.

Physics
1 answer:
Marysya12 [62]2 years ago
6 0

Answer:

Force applied

Explanation:

An object will remain at its state of rest unless a non zero for act on it

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Please help me with part b.
Lerok [7]
I'm sorry but I don't see anything that says part B
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8 0
3 years ago
When light passes the
Leno4ka [110]

When light crosses the boundary between layers with different densities, the light is refracted. (A).

3 0
3 years ago
An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou
frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is u=3\ m/s

speed of camera u_1=20\ m/s

Initial separation between camera and balloon is d_o=5\ m

Suppose after t sec of  throw camera reach balloon then,

distance travel by balloon is

s=ut

s=3\times t

and distance travel by camera to reach balloon is

s_1=ut+\frac{1}{2}at^2

s_1=20\times t-\frac{1}{2}gt^2

Now

\Rightarrow s_1=5+s

\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t

\Rightarrow 5t^2-17t+5=0

\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}

\Rightarrow t=\dfrac{17\pm 13.747}{10}

\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is  t=0.3253\ s

velocity is given by

v=u+at

v=20-10\times 0.3253

v=16.747\ m/s

and position of camera is same as of balloon so

Position is =5+3\times 0.3253

=5.975\approx 6\ m

8 0
3 years ago
Initially, a particle is moving at 5.25 m/s at an angle of 35.5° above the horizontal. Three seconds later, its velocity is 6.0
ivolga24 [154]

Answer:

 a =( -0.32 i ^ - 2,697 j ^)  m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

  v₀ₓ = v₀ cos θ

  v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

   v_{oy} = v₀ sin θ

 v_{oy}= 5.25 sin35.5

v_{oy} = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

v_{y} = v₀ sin θ

v_{y} = 6.03 sin (-56.7)

v_{y} = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

    a = (v_{f} - v₀) /t

    aₓ = (3.31 -4.27)/3

    aₓ = -0.32 m/s²

    a_{y} = (-5.04-3.05)/3

   a_{y} =  -2.697 m/s²

6 0
3 years ago
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
3 years ago
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