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Llana [10]
3 years ago
8

How is net force calculated when two forces act on an object in opposite directions?

Physics
1 answer:
Lyrx [107]3 years ago
6 0
Force will push against each other like opposite magnets
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If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
kondaur [170]
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
3 0
3 years ago
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

4 0
3 years ago
Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
A train travels 77 kilometers in 1 hour, and the 66 kilometers in 1 hour. What is the average speed?
Kipish [7]

Average speed = (total distance covered) / (time to cover the distance)

Total distance = (77km + 66km) = 143 kilometers

Time to cover the distance = 2 hours

Average speed = (143 km) / (2 hours) =  71.5 km per hour
6 0
3 years ago
A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light
MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

        n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

        v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

        v = d / t

        t = d / v

        t = d n / c

Let's look for the times on each sheet

Ice

        t₁ = 1.4 10⁻² 1.31 / 3 10⁸

        t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

        t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

        t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

         t = t₁ + t₂

         t = (0.6113 + 1.52) 10⁻¹⁰

         t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

        d = c t

       d = 3 10⁸ 2.13 10⁻¹⁰

       d = 6.39 10⁻² m

       d = 6.39 cm

3 0
3 years ago
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