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2 years ago

10

A ball is dropped from a height of 1m. If the coefficient of restitution between the ball and the surface is 0.6, what is the he

ight the ball rebounds to?

1 answer:

gtnhenbr [62]2 years ago

7 0

Answer:

0.7

Explanation:

1 + 0.6 = 0.7 I think

hope this helps

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if a gas has a gage pressure of 156 kpa its absolute pressure is approximately .....A... 56 kPa ....B....100 kPa.......C......25

NeX [460]

256 kPa because p-guage + p-absolute + p-atmospheric = 256

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3 years ago

A small box slides down a ramp on a friction with surface. If the total energy of the system is conserved, which computational m

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3 years ago

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

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3 years ago

If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he disting

iVinArrow [24]

Answer:

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

8 0

3 years ago

You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward

pickupchik [31]

Answer:

Displacement is 565.69 m at 45° west of north

Explanation:

Let north represent positive y axis and east represent positive x axis.

We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.

3 blocks toward north = 300 j m

4 blocks toward west = -400 i m

1 blocks toward north = 100 j m

Total displacement = -400 i + 400 j m

Magnitude

$s=\sqrt{(-400)^2+400^2}=565.69m$

Direction,

$\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0$

Direction is 45° west of north.

Displacement is 565.69 m at 45° west of north

4 0

3 years ago