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jek_recluse [69]
3 years ago
10

A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)

Physics
1 answer:
Tasya [4]3 years ago
7 0

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

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B- the acceleration is greater for the more massive rock
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2 years ago
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A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh
MrMuchimi

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

W=\vec F \cdot \vec r

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

W=F.d

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

W=350\ N\ .\ 2\ m=700\ Joule

4 0
3 years ago
A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car
jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0
2 years ago
A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

6 0
3 years ago
A cube of iron (Cp = 0.450 J/g•°C) with a mass of 55.8 g is heated from 25.0°C to 49.0°C. How much heat is required for this pro
Rus_ich [418]
Q = mcΔT 
<span>q = 55.8g x 0.450J/gC x 23.5C </span>
<span>q = 590. J ................ to three significant digits

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

</span>
8 0
3 years ago
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