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Sliva [168]
3 years ago
10

In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:

Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

−x direction decreasing in speed.

Explanation:

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Why are images reflected from a rough surface not as clear as those reflected from a smooth
ioda

Answer:

Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

Explanation:

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7 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
In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff
Fynjy0 [20]

Answer:

1.25 m

Explanation:

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

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

8 0
3 years ago
Bob walks 370 m south, then jogs 410 m southwest, then walks 370 m in a direction 28 degrees east of north.
wlad13 [49]
We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y

Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m

displacement = √(116² + 333²)
= 353 m

Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
5 0
3 years ago
Read 2 more answers
One watt is wqual to 1kg .m/s³​
djverab [1.8K]

1 watt = 1 joule per second = 1 newton meter per second = 1 kg m2 s-3

6 0
3 years ago
Read 2 more answers
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