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gayaneshka [121]
3 years ago
10

How are credit unions different from banks?

Physics
1 answer:
aleksandrvk [35]3 years ago
7 0
Answer: B ≈ Credit unions are owned by stockholders rather than partners
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Match the player positions with his or her job on the court.
velikii [3]
2.c 
3.b
1.a
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4 0
3 years ago
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Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm
lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

4 0
3 years ago
A tennis ball is thrown against a vertical concrete wall that is fixed to the ground. The ball bounces off the wall. How does th
cestrela7 [59]

Answer:

Explanation:

The forces compare together as a result of the fact that the force exerted by that of the ball and the force exerted by that of the wall both have the same magnitude.

8 0
3 years ago
A hot ball of gas in outer space that creates its own energy is called a _____.
Fynjy0 [20]
It's called a star when it creates its own energy
3 0
3 years ago
A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
3 years ago
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