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Misha Larkins [42]
3 years ago
10

Astronauts who traveled to the moon were able to float slightly as they walked along its surface. Why were the astronauts able t

o do this? The weight of the astronaut on Earth is less than his weight on the moon. The mass of the astronaut on Earth is less than his mass on the moon. The weight of the astronaut on Earth is greater than his weight on the moon. The mass of the astronaut on Earth is greater than his mass on the moon.
Physics
2 answers:
Jet001 [13]3 years ago
5 0

Answer:The mass of the astronaut on Earth is greater than his mass on the moon

Explanation:

Astronauts who traveled to the moon were able to float slightly as they walked along its surface. Why were the astronauts able to do this? The weight of the astronaut on Earth is less than his weight on the moon. The mass of the astronaut on Earth is less than his mass on the moon. The weight of the astronaut on Earth is greater than his weight on the moon. The mass of the astronaut on Earth is greater than his mass on the moon.

timurjin [86]3 years ago
3 0

Answer: The weight of the astronaut on earth is greater than his weight on the moon.

Explanation:

The reason I got this answer is because the astronaut will have the same mass on the moon as he does on earth. Also the weight of a person or object is determined by gravity. So since the moons gravity isn’t as strong as earths gravity the astronauts are lighter in terms of weight, and this is how they slightly float.

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You have three identical metallic spheres, A, B and C, fixed to isolating pedestals. They all start off uncharged. You then char
kaheart [24]

Answer:

Charge on B is 12 uC.

Explanation:

Initial charge on A = 32 uC

Initial charge on B and C = 0

Now A touches to B, so the charge on A and B both is

q = (32 + 0) / 2 = 16 uC

Now A touches to C, so the charge on A and C both is

q' = (16 + 0) / 2 = 8 uC

Now again A touches to B so the charge on B is

q''= (8 + 16) / 2 = 12 uC  

7 0
3 years ago
An object has 16 N of force being applied to the right, 16 N of force being applied to the left, and 6 N of force being applied
umka2103 [35]

Answer:

the net force on the object is 6 N.

Explanation:

Given;

force applied to the right on the object, F_r = 16 N

force applied to the left of the object, F_l = 16 N

force applied downward on the object, F_d = 6 N

The net horizontal force on the object = F_r - F_l = 16N -16N = 0

The net vertical force on the object = 6 \ N \ downward

The net force on the object is calculated as;

F_{net}^2 = 6^2 + 0\\\\F_{net}^2 = 36\\\\F_{net} = \sqrt{36} \\\\F_{net} = 6 \ N

Therefore, the net force on the object is 6 N.

7 0
3 years ago
A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at
stira [4]

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

\omega =\dfrac{2\pi N}{60}

\omega =\dfrac{2\pi \times 150}{60}

ω = 15.7 rad/s

The distance of center of mass from 150 g

r=\dfrac{150\times 0+240\times 33}{150+240}\ cm

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

6 0
3 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
The distance covered by a car at a time, t is given by x = 20t + 6t4, calculate
Anni [7]

Answer:

(i) v = 44 m/s

(ii) a = 72 m/s^2

Explanation:

You have the following equation for the potion of a car:

x=20t+6t^4

(i) The instantaneous velocity is the derivative of x in time:

\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3

for t = 1 is:

v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s

(ii) The instantaneous acceleration is the derivative of the velocity:

\frac{dv}{dt}=24(3)t^2=72t^2

for t = 1

a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2

8 0
4 years ago
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