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Lemur [1.5K]
2 years ago
11

Two students push on a 5-kg cart from opposite sides

Physics
2 answers:
icang [17]2 years ago
6 0
It will break as they are pushing from both sides
sweet-ann [11.9K]2 years ago
4 0

Answer:

nothing will happen the cart will be broken or as it is

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aliina [53]

in the same direction as the displacement vector and the motion

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3 years ago
C. is what percent of 125?​
solmaris [256]

Answer:

Step 1: We make the assumption that 125 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$.

Step 3: From step 1, it follows that $100\%=125$.

Step 4: In the same vein, $x\%=125$.

Step 5: This gives us a pair of simple equations:

$100\%=125(1)$.

Explanation:

4 0
3 years ago
A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0
ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

c_{a} = specific heat of aluminium = 900

T_{ai} = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

c_{w} = specific heat of water = 4186

T_{wi} = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M c_{a} (T_{ai} - T) = m c_{w} (T - T_{wi} )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0
3 years ago
How can the difference in the brightness of spectral lines be explained?
Ksenya-84 [330]
<span>more lines = a lot of electrons returning back to ground state from same level</span>
5 0
3 years ago
PLEASE HELP 15 POINTS AND BRAINIEST!!!<br> Reporting fake answers
Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

T = 2\pi \sqrt{\frac{r^3}{GM} }

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

 

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N

5 0
2 years ago
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