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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

A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca

Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

$\theta$ = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

$P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J$

Work done is 92704.5 J

Force required is given by

$F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N$

The force is 596.44737 N

7 0

3 years ago

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago

An amplifier has a 50 watt output and a 5 watt input. what is the gain in decibels for this amplifier, rounded to the nearest de

Marrrta [24]

Gain in decibels is given by;

Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input

Substituting;

Gain in db = 10 * log (50/5) = 10 db

6 0

3 years ago

Physics question plz help

blagie [28]

Answer:

B. $30\,s\,< t \le 40\,s$ .

Explanation:

The vehicle is travelling eastwards when its velocity is positive and westwards when velocity is negative. According to the graph, $v > 0$ for $30\,s\,< t \le 40\,s$ . Hence, correct answer is B.

5 0

3 years ago