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photoshop1234 [79]
3 years ago
14

okay so there is a wheel and it shows it when its high, in the middle, and below. my question is when is its potenial energy the

greatest?
Physics
1 answer:
earnstyle [38]3 years ago
4 0
At the highest point 


cheers

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object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Cal
VARVARA [1.3K]

The distance between object P1 and its image formed is determined as 36 m.

<h3>Distance of the image</h3>

The distance of the image formed by object P1 is calculated as follows;

In a plane mirror; object distance = image distance

image distance of P1 = 18 m

distance between object and image = 18m + 18 m = 36 m

Thus, the distance between object P1 and its image formed is determined as 36 m.

Learn more about plane mirrors here: brainly.com/question/1126858

#SPJ1

8 0
2 years ago
Which of the following statements would be classified as an aphorism?
Veseljchak [2.6K]
A bird in the hand is worth two in the bush
or
actions speak louder than words
or
her bark is worse than her bite
i hope this helps these are examples of aphorism
5 0
3 years ago
An object has an angular velocity. It also has an angular acceleration due to torques that are present. Therefore, the angular v
shusha [124]

Answer:

 α = 0 ,    w = w₀

Explanation:

Torque is related to angular acceleration by Newton's second law for rotational motion.

        τ = I α

Where τ is the torque, I the moment of inertia and α the angular acceleration.

If we apply an external torque for the sum of all torques to be zero, the angular acceleration must fall to zero

         α = 0

 

Since the acceleration is zero, the angular velocity you have at that time is constantly killed.

           w = w₀ + α t

          w = w₀ + 0

8 0
3 years ago
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the
marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball u=10\ m/s

height of window h=20\ m

Using Equation of motion

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

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

Y=ut+\frac{1}{2}at^2+20

Y=10\times t+0.5\times (-9.8)t^2+20

Y=-4.9t^2+10t+20

(b)highest point is obtained at v=0

v^2-u^2=2as

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)-10^2=2\times (-9.8)\times s

s=\frac{100}{19.6}

s=5.102\ m

Highest Point will be s+20=25.102\ m

(c)Time taken when the ball hit the ground i.e. at Y=0

-4.9t^2+10t+20=0

t=3.28\ s

impact velocity v=\sqrt{2\times 9.8\times 25.102}

v=22.181\ m/s

7 0
3 years ago
Which description matches with which graph?
VikaD [51]

Answer:

Description 1 matches C.

Description 2 matches A.

Description 3 matches D.

Description 4 matches B.

Explanation:

Description 1 matches C. From graph C, we can see that the <em>distance remains constant at a particular value</em>. Hence, the car is <em>not traveling</em> and hence it is <em>stopped.</em>

Description 2 matches A. <em>Speed is determined by the gradient of a distance-time graph</em>. A depicts a <em>linear graph</em> which tells us that the <em>gradient is constant </em>and hence the car is at <em>constant speed</em>. The fact that the<em> distance is increasing</em> shows that the<em> car is moving forward</em>.

Description 3 matches D. Graph of D becomes <em>less steeper as time progresses</em>. This tells us that the <em>gradient of the graph is decreasing </em>and hence the <em>speed of the car is decreasing.</em>

<em />

Description 4 matches B. Graph B shows <em>distance being decreased.</em> This tells us that the <em>car is coming back</em>.

4 0
3 years ago
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