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Illusion [34]
2 years ago
9

Which objects have both potential and kinetic energy? Check all that apply.

Physics
2 answers:
wariber [46]2 years ago
3 0
A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.
denis23 [38]2 years ago
3 0

Answer:

a sky diver who is midway through his descent to the ground

a roller coaster car halfway down the second hill

a falling raindrop

Explanation:

From the law of conservation of energy, it can neither be created nor destroyed but can be converted from one form to another.

The sum of kinetic energy and potential energy is mechanical energy. A body has potential energy due to its position or configuration and kinetic energy due to its speed.

A body can have both potential energy and kinetic energy.

The correct options are:

  • a sky diver who is midway through his descent to the ground
  • a roller coaster car halfway down the second hill
  • a falling raindrop

A boulder resting on top of a mountain  has just potential energy.

A parked car  also has just potential energy.

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Traffic officials indicate, it takes longer to ______ when you drive fast.
nignag [31]
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
7 0
2 years ago
Read 2 more answers
PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0
2 years ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, M=75\ kg

Mass of the ball is, m=4\ kg

Initial velocity of the man is, u_m=0\ m/s(rest)

Initial velocity of the ball is, u_b=0\ m/s(rest)

Final velocity of the ball is, v_b=3.50\ m/s

Final velocity of the man is, v_m=?\ m/s

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of man and ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Now, initial momentum = final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0
2 years ago
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Airida [17]

Answer:

See explanation

Explanation:

Solution:-

- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.

- The angle that each fringe forms on the screen is defined by ( θ ).

- The order of bright/dark spot is defined by an integer ( n )

- The wavelength of the incident light is ( λ )

- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The above given formulation is for constructive interference.

- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.

- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Substitute the above approximation in the given relation of diffraction gratings as follows:

                            y_n = n*lamda*L*N

- To double the distance between the screen and grating we will use the above relation with ( 2L ):

                            yn ∝ L

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.

- Similarly, the line density of the grating ( N ) was doubled. Then,

                            yn ∝ N

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.

4 0
2 years ago
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s
Anastaziya [24]

Answer:

the correct answer is c     v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

            v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

           a₁ = v² / 2x

           a₁ = 25² / (2 120)

           a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

           v₁² = 0 + 2 a1 x₂

           v₁ = Ra (2 2,6 60)

           v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c     v₁> 12.5 m / s

5 0
2 years ago
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