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jeyben [28]
3 years ago
14

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

y experiencing?
A. tension
B. external
C. internal
D. compression
Physics
2 answers:
torisob [31]3 years ago
7 0
When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth, the type of force they are experiencing is <span>A. tension. Tension is the opposite of compression which is pulling of the astronaut from the ground or Earth</span>
Anvisha [2.4K]3 years ago
3 0

Answer:

A. tension

Explanation:

The reason that astronauts float is due to their free fall, that is, when astronauts are in Earth orbit, Earth gravity attracts them. when astronauts travel to the moon, this attraction that the earth imposes on them is much smaller, called the outer attraction. This external attraction exerts on the astronauts a tention, since the gravity of the moon is insignificant.

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A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s
irga5000 [103]
14 m/s in the direction of the truck
5 0
3 years ago
An increase in the magnitude of velocity is??
bazaltina [42]
It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
hence
there is a force acting on the object, in the direction of the velocity
8 0
3 years ago
On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m

From Hooke's law

mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0
3 years ago
2NaOH+H2SO4 Na2SO4+H2O the equation is balanced/unbalanced because the number of hydrogen atoms and oxygen/sodium is equal/not e
RUDIKE [14]
Your equation is:
2NaOH + H_{2}SO_{4}  ==\ \textgreater \  Na_{2}SO_{4} + H_{2}O

An equation is balanced only if there are the same number of atoms of each element on both sides of the arrow - aka same number of atoms of each element in both reactants (left of the arrow) and products (right of the arrow). 

It'll be easiest to tackle this by counting up the number of atoms of each element on the left and on the right and comparing those numbers. If there is a number in front of the entire compound, that means that number applies to all elements in the compound. If the number is a subscript (little number to the right of the element), that means that number only applies to the element that the subscript is attached to:
1) On the left, you have:
2NaOH + H_{2}SO_{4}\\&#10;\\&#10;Na:  \: 2 \: atoms\\&#10;O: \: 2 + 4 = 6 \: atoms\\&#10;H: \: 2 + 2 = 4 \: atoms\\&#10;S:  \: 1 \: atom

2) On the right, you have:
Na_{2}SO_{4} + H_{2}O\\&#10;\\ &#10;Na: \: 2 \: atoms\\ &#10;O: \: 4 + 1 = 5 \: atoms\\ &#10;H: \: 2 \: atoms\\ &#10;S: \: 1 \: atom

You can see that the number of oxygen and hydrogen atoms aren't equal on both the left (reactants) and the right (products), so the equation is unbalanced.

Your final answer is "T<span>he equation is unbalanced because the number of hydrogen atoms and oxygen is not equal in the reactants and in the products."</span>


5 0
3 years ago
Read 2 more answers
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

3 0
3 years ago
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