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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

8_murik_8 [283]

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.

6 0

3 years ago

Potential energy is the energy

alina1380 [7]

....possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors

7 0

4 years ago

Read 2 more answers

Suppose a certain battery has an internal emf of 9.00 V but the potential difference across its terminals is only 80.0 % of that

Rudiy27

Answer:

$1140.48\times 10^{-6}J$

Explanation:

We have given that the battery has an internal emf of 9 volt

So E = 9 volt

Capacitance $C=44\mu F=44\times 10^{-6}F$

It is given that on the terminal voltage is only 80% of potential difference

So V = 0.8×9 = 7.2 volt

We know that energy stored in the capacitor is given by

$E=\frac{1}{2}CV^2=\frac{1}{2}\times 44\times 10^{-6}\times 7.2^2=1140.48\times 10^{-6}J$

8 0

3 years ago

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat

agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2 f^2}$

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0

3 years ago

The acceleration due to gravity at the surface of a planet depends on the planet's mass and size; therefore other planets will h

Ahat [919]

Answer: $1.88 m/s^2$

Explanation:

Given

initial horizontal speed of =6.55 m/s

height of rock=1.4 m

Horizontal distance=8 m

Time to travel 8 m

$8=6.55\times t$

$t=\frac{8}{6.55}=1.22 s$

Time to cover 1.4 m

$h=\frac{at^2}{2}$

$1.4=\frac{g'1.22^2}{2}$

$g'=1.88 m/s^2$

6 0

3 years ago