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guajiro [1.7K]
3 years ago
8

How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1

.6N/kg.
Physics
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

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DaniilM [7]

Answer:

Fx = 35.36 N

Fy = 35.36 N

Explanation:

From the question,

The X component of the force is

Fx = Fcos∅.................. Equation 1

Where Fx = X component of the force, F = Force, ∅ = Angle to the horizontal.

Give: F = 50 N, ∅ = 45°

Substitute into equation 1

Fx = 50(cos45°)

Fx = 50(0.7071)

Fx = 35.36 N

Similarly,

For Y component

Fy = Fsin∅

Where F y = Y component

Fy = 50(sin45°)

Fy = 50(0.7071)

Fy = 35.36 N

4 0
3 years ago
A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
Doss [256]

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

5 0
3 years ago
A flowerpot falls off a balcony 85m above the street how long does it take to hit the ground
GenaCL600 [577]
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Read 2 more answers
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
4 years ago
What is conclusion about potential difference (voltmeter readings) in a parallel electric circuit ?
Ainat [17]

Answer: the conclusion is that

Explanation:

5 0
3 years ago
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