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3 years ago

Pressure __________ with depth to support the fluid weight above.

Physics

1 answer:

Neporo4naja [7]3 years ago

4 0

Answer:

B. Increases, also B is before C

Explanation:

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IS IT B!! if not pls helpp!

natita [175]

Answer:

I believe you are correct but we just started this unit

Explanation:

6 0

3 years ago

Classify the following situations as involving balanced or unbalanced forces.

tester [92]

Answer:

1) Unbalanced

2) Balance

3) Balanced

4) Unbalanced

5) Unbalanced

Explanation:

For 1 and 5, since the objects are not at a constant speed/velocity, their forces must be unbalanced.

For 2 and 3, their speeds are constant so that means the force is balanced. If the bycicle or box started accelerating, that would indicate an unbalanced force

For 4, the speed is constant, but it's direction is not indicating another unbalanced outside force causing it to turn. The force is unbalanced.

5 0

3 years ago

G = 10 N/kg or 10 m/s2

Irina18 [472]

Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

a) The initial potential energy is:

$U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)$

$U_{g} = 40\,J$

b) The efficiency of the bounce is:

$\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%$

$\eta = 70\,\%$

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

$K = U_{g}$

$U_{g} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

$v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }$

$v = 20\,\frac{m}{s}$

5 0

3 years ago

At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2×104−kg truck going 20 km/h?

Ostrovityanka [42]

<u>Answer:</u>

950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

<u>Explanation:</u>

Kinetic energy of body with mass m and moving at velocity v is given by

$KE=\frac{1}{2} mv^2$

KE of truck $= \frac{1}{2}*3.2*10^4*20^2$

KE of subcompact car $= \frac{1}{2}*950*v^2$

Equating

$\frac{1}{2}*3.2*10^4*20^2= \frac{1}{2}*950*v^2\\ \\ v=116.08 km/hr$

So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

6 0

4 years ago

A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains

makkiz [27]

Answer:

3.834 m/s

Explanation:

h = 0.750 m

vx = 5 m/s

Let the initial vertical velocity is vy.

Final vertical velocity is zero

use third equation of motion along y axis

v^2 = vy^2 - 2 x g x h

0 = vy^2 - 2 x 9.8 x 0.75

vy^2 = 14.7

vy = 3.834 m/s

5 0

4 years ago