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2 years ago

EARTH SCIENCE HELP FAST PLEASE

Physics

1 answer:

xxTIMURxx [149]2 years ago

8 0

I think its C not sure tho

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How much gravitational potential energy does a 1.0 kg hammer have on a shelf 1.5 m above the ground?

jok3333 [9.3K]

Answer:

14.7 J

Explanation:

PE=MGH

PE= 1.0 x 9.8 x 1.5 = 14.7 J

8 0

2 years ago

If the mass of a body is 9.8kg on the earth, what would be its mass on the moon?

devlian [24]

Answer:

it will remain same because mass of the body is constant everywhere

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2 years ago

How can accuracy be limited?

lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0

2 years ago

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of

Anestetic [448]

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

$KE = \frac{1}{2}m* v^{2}$

It is given:

v = 8.5 m/s

m = 91 kg

So:

$KE= \frac{1}{2}*91*8.5^{2} =3,287.4J$

b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height

Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

3 0

2 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

2 years ago