What type of energy does mechanical energy turn into when friction is acting

A ball accelerates from rest down a ramp at 2.4 m/s^2. Write an equation that could be used to determine the balls finals positi

Alik [6]

Answer:

x=2.4t+4.9t^2

Explanation:

This equation is one of the kinematic equations to solve for distance. The original equation is as follows:

X=Xo+Vt+1/2at^2

We know that the ball starts at rest meaning that its initial velocity and position is zero.

X=0+Vt+1/2at^2

Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.

X=Vt+4.9t^2

Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.

Now, substitute in your velocity value.

X=2.4t+4.9t^2

8 0

2 years ago

A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b

mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

$V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s$

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0

3 years ago

A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

lakkis [162]

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

6 0

3 years ago

What happens when sound waves move faster?

Lady bird [3.3K]

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6 0

2 years ago

PLEASE HELP

Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0

2 years ago