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3 years ago

A rolling ball has 3,276) of energy, what is it's mass when it is rolling at a velocity of 53m/s? what is the mass

Physics

1 answer:

hram777 [196]3 years ago

3 0

Initial velocity U = 0

Final velocity V = 32.23 m/s

Explanation:

Given that a coin is dropped from the top of the Tower of Pisa, 53m above the ground.

What is the coin's initial velocity ?

Since the coin is dropped from the tower, the initial velocity U will be equal to zero.

Therefore, U = 0

But the final velocity V will be calculated by using the formula

V^2 = U^2 + 2gH

V^2 = 0 + 2 × 9.8 × 53

V^2 = 1038.8

V = sqrt ( 1038.8)

V = 32.23 m/s

sorry if did not help :(

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You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an

Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

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3 years ago

Contrast the image formed by a convex lens when an object is located more than twice the focal length from the lens with the ima

matrenka [14]

Answer:

$\begin{array}{lll}&\underline{Object \ at \ more \ than \ 2\times Focus} & \underline{Object \ at \ less \ than \ Focus}\\\\1. \ Location \ of \ image &Same \ side \ as \ object&Opposite \ side \ of \ lens\\\\2. \ Orientation \ of \ image &Inverted&Upright\\\\3. \ Type \ of \ image&Real&Virtual\\\\4. \ Size\ of \ image&Smaller \ than \ object& Larger \ than \ object\end{array}$

Explanation:

The location, orientation, size, and type of image formed by a convex lens are related to the position of the image location in front of the lens

Object >2·F = The image formed by a convex lens when the object is located more than twice the focal length from the lens

Object < F = The image formed by the convex lens when the object is located between the lens and the focal length

$\begin{array}{lll}&\underline{Object \ > \ 2\cdot F} & \underline{Object < F}\\\\1. \ Location \ of \ image &Same \ side \ as \ object&Opposite \ side \ of \ lens\\\\2. \ Orientation \ of \ image &Inverted&Upright\\\\3. \ Type \ of \ image&Real&Virtual\\\\4. \ Size\ of \ image&Smaller \ than \ object& Larger \ than \ object\end{array}$

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3 years ago

A racing car travels with a constant tangential speed of 80.8 m/s around a circular track of radius 602 m. (a) Find the magnitud

KatRina [158]

Answer:

10.84 m/s2 radially inward

Explanation:

As the car is traveling an a constant tangential speed of 80.8 m/s, the total acceleration only consists of the centripetal acceleration and no linear acceleration. The formula for centripetal acceleration with respect to tangential speed v = 80.8 m/s and radius r =602 m is

$a = \frac{v^2}{r} = \frac{80.8^2}{602} = 10.84 m/s^2$

b) The direction of this centripetal acceleration is radially inward

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4 years ago

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avanturin [10]

Answer:

not sure

Explanation:

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3 years ago

Sue played four games of golf the modal score was 98 and the mean 100 and the range 10 What were the scores for the games

Lyrx [107]

Answer:

Explanation:

Let the highest and lowest score be a and b respectively .

a - b = 10

a = 10 + b

The numbers in descending order are a , 98 , 98 , b .

mean = 100 , so

4 x 100 = a + 98 + 98 + b = 10 + b + 98 + 98 + b = 206 + 2b

400 = 206 + 2b

2b = 194

b = 97

a = 107

So the scores were as follows

107 , 98 , 98 , 97

7 0

3 years ago

A rolling ball has 3,276) of energy, what is it's mass when it is rolling at a velocity of 53m/s? what is the mass​

A rolling ball has 3,276) of energy, what is it's mass when it is rolling at a velocity of 53m/s? what is the mass