Ask question

Ask question

All categories

3 years ago

5

When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth. Which type of pull are the

y experiencing?

2 answers:

Georgia [21]3 years ago

5 0

Answer:

The pull the astronauts feel in the moon will be gravitational too.

Explanation:

The pull will be gravitational too because there's a gravitational force exerted by the moon over the astronaut, this force keeps pulling the astronaut towards the moon surface but it's weaker than in earth because the mass of the moon is significantly less that the mass of the earth.

postnew [5]3 years ago

3 0

They are experiencing the pull of leaving the atmosphere

You might be interested in

You connect five identical resistors in series to a battery whose EMF is 12.0 V and whose internal resistance is negligible. You

neonofarm [45]

Answer:

Explanation:

Total resistance in the circuit

= EMF / current in the circuit

= 12 / .969

= 12.383 ohm

This resistance consists of 5 identical resistances in series

resistance of each resistor

= 12.383 / 5

= 2.476 ohm .

potential difference on each

= current x resistance of each

= .969 x 2.476

= 2.399 V

= 2.4 V

4 0

3 years ago

A force of 10 n acts on a mass of 5 kg producing an acceleration of 1.5 m/s2 what is the magnitude of the force of friction acti

prisoha [69]

The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2

8 0

3 years ago

The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

3 0

3 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$

$m_1u_1 +m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

$m_2u_2 = (m_1+m_2)v_f$

Rearranging to find the final velocity

$v_f = \frac{m_2u_2}{ (m_1+m_2)}$

$v_f = \frac{ 12500*7.8}{ 12500+7430}$

$v_f = 4.8921ft/s$

The expression for the impulse received by the first car is

$I = m_1 (v-u)$

$I = \frac{W}{g} (v-u)$

Replacing,

$I = \frac{12500}{32.2}(4.89-7.8)$

$I = -1129.65lb\cdot s$

The negative sign show the opposite direction.

7 0

3 years ago

What types of electromagnetic radiation has a shorter wavelength than ultraviolet?

Lunna [17]

X-rays and gamma rays are the only electromagnetci waves with a shorter wavelength, gamma rays being the smallest. Hope this helps ;)

4 0

3 years ago