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andrew-mc [135]
2 years ago
6

A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What

must be true of the wheelbarrow's motion?
Physics
2 answers:
Tamiku [17]2 years ago
7 0

Answer:

its acceleration must be 0.

Explanation:

According the second Newton's law, the net force applied in a object is equal to the mass of the object multiplied by its acceleration:

F_{net} =m.a

In this case, the net force is

F_{net}=5N-5N=0

What makes sense since both forces have the same value and opposite directions.

Since the mass of the object can't be zero, the acceleration is the one that must be 0 in order fulfill the equation of the second newton's law.

Another option is that, since both forces are opposite, the wheelbarrow could be at rest. Though this is correct, this is not always true.

In conclusion, the only  statement that always is true about the wheelbarrow's motion is that its acceleration must be 0.

mezya [45]2 years ago
4 0
Assuming no other forces are acting on the wheelbarrow, it must be stationary.
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A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
3 years ago
Read 2 more answers
Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini
kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

  • Mass of cart one M1  = 150kg
  • Initial Velocity U1 = 8m/s
  • Final VelocityV1 = 5 m/s

Mass of cart two M2  = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0
2 years ago
13
gladu [14]

v = u + at

50 = 0 + a*10

50 = 10a

a = 5 m/s^2

4 0
2 years ago
A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I
Mashutka [201]

Answer:

the water level remains same

Explanation:

This can be explained by Archimedes's principle which says that the wood will sink if weight of wood is more than the weight of the water displaced with weight equal to the water displaced otherwise the wood will float.

Therefore, buoyancy or the buoyant force is the same as the weight of wood, the weight of the water displaced by wood is also the same as that of the weight of wood.

Thus, we can see that the weight of the wood remains same and so is the level of water.

6 0
2 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
3 years ago
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