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lianna [129]
3 years ago
14

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
2 answers:
wlad13 [49]3 years ago
7 0
<span>potential energy of a swing is greatest at the top as height is directly proportional to the potential energy 

In short, Your Answer would be Option A

Hope this helps!</span>
OLga [1]3 years ago
5 0
At the top, due to the fact that POTENTIAL energy is Stored energy.
You might be interested in
Guys No one's answering my question so sad! Once again I'm asking the same question –Here
nexus9112 [7]

For the front glass of the car to get wet, V_c \geq 10 \ m/s.

The given parameters:

  • <em>Speed of the car, = Vc</em>
  • <em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

V_{C/R} = V_C- V_R

  • If the speed of the car equals the speed of the rain, the rain will fall behind the car.
  • If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
  • If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, V_c \geq 10 \ m/s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0
3 years ago
The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
3 years ago
In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hits an obstacle and moves in th
qwelly [4]
Change in velocity = Final velocity - Initial velocity

Initial velocity = 20 m/s
Final velocity = 10 m/s

Change in velocity = 10 - 20 = -10m/s

Correct option B

Hope This Helps You!
4 0
4 years ago
A basketball player throws a basketball m = 1 kg straight up with an initial speed of v0 = 9.5 m/s. The ball leaves his hand at
brilliants [131]

To solve this problem we will apply the concepts related to energy conservation. So that the initial energy on the system is equivalent to the final energy.

The initial or final energy will also be the TOTAL mechanical energy of the body.

In the case of the initial energy we will have two types of energy on the body: Kinetic energy and potential energy.

For the case of the final energy we will only have the potential energy in terms of the height h_m, the mass m, and the gravity g

E_i = E_f

KE_i + PE_i = PE_f

\frac{1}{2} mv_0^2 +mgh_0 = mgh_m

E = mgh_m

The total mechanical energy will be equivalent in the terms required, to the final potential energy.

6 0
3 years ago
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00
ANTONII [103]

Answer:

a) 200m, 100m/s

b) 710.20m

c) -117.98 m/s

d) 26.24 s

Explanation:

To solve this  we have to use the formulas corresponding to a uniformly accelerated motion problem:

V=Vo+a*t (1)

X=Xo+Vo*t+\frac{1}{2}*a*t^2\\ (2)

V^2=Vo^2+2*a*X (3)

where:

Vo is initial velocity

Xo=intial position

V=final velocity

X=displacement

a)

X=0+0*4+\frac{1}{2}*25*4^2

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.

X=200m

V=0+25*4

V=100m/s

b)

The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8\frac{m}{s^2} because it is going downward.

0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m

c)

In order to find the velocity when it crashes, we can use the formula (3).

the initial velocity is 0 because in that moment is starting to fall.

V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s

the minus sign means that the object is going down.

d)

We can find the total amount of time adding the first 4 second and the time it takes to going down.

to calculate the time we can use the formula (2) setting the reference at 200m:

-200=0+100*t+\frac{1}{2}*(-9.8)*t^2

solving this we have: time taken= 22.24 seconds

total time is:

total=22.24+4=26.24 seconds.

3 0
3 years ago
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