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3 years ago

How do you find the net force acting on an object?

1 answer:

6 0

It's D. By "net" they mean the overall force the object experiences, so sum all the force vectors, those in a negative direction (eg friction) should be subtracted.

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Low-frequency red light is used in dark rooms because it does not expose the film. This is because the red light has a low

Alex

energy. It will not disrupt the picture developing process by overexposing too much light on the film.

4 0

3 years ago

Read 2 more answers

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

3 years ago

(Mass vs. Weight) HELP PLZ!!

Lina20 [59]

$\huge \boxed { \sf{Answers}}$

c. The weight of an object on the moon will be the same as its weight on Earth. It is false because the weight of an on the moon will be 1/6 th times its weight on Earth.
d. The weight of an object is its mass multiplied by the force of gravity. The statement is false because the formula of weight is mass × acceleration due to gravity, not force of gravity.
e. The mass and weight of an object are the same thing. The statement is false because mass means a body of matter. While weight of an object is its mass multiplied by the force of gravity.
f. The mass of an object is the force of gravity acting upon an object. It is false because it will be the weight of the object not mass.
So, the answers are c, d, e and f.

Hope you could understand.

If you have any query, feel free to ask.

7 0

3 years ago

Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the

Hitman42 [59]

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

a_a = a_b / 3

5 0

3 years ago

A pressure sensor inside of a mixing tank is designed to turn red when the pressure inside the tank exceeds 1.9 kPa. If the sens

spin [16.1K]

Answer:

19 N

Explanation:

From the question given above, the following data were obtained:

Pressure (P) = 1.9 kPa

Length (L) = 10 cm

Force (F) =?

Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:

1 KPa = 1000 N/m²

Therefore,

1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa

1.9 KPa = 1900 N/m²

Thus, 1.9 KPa is equivalent to 1900 N/m².

Next, we shall convert 10 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

10 cm = 10 cm × 1 m / 100 cm

10 cm = 0.1 m

Thus, 10 cm is equivalent to 0.1 m

Next, we shall determine the area of the square. This can be obtained as follow:

Length (L) = 0.1 m

Area of square (A) =?

A = L²

A = 0.1²

A = 0.01 m²

Thus, the area of the square is 0.01 m².

Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:

Pressure (P) = 1900 N/m²

Area (A) = 0.01 m²

Force (F) =?

P = F/A

1900 = F / 0.01

Cross multiply

F = 1900 × 0.01

F = 19 N

Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.

3 0

3 years ago