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Romashka-Z-Leto [24]
3 years ago
6

An aluminum bar has a mass of 9 kg in the air. Calculate its volume. Now imagine that you submerge the aluminum bar in water han

ging it from a rope. Calculate the apparent weight it would have in the water.
Physics
1 answer:
yan [13]3 years ago
7 0

Answer:

55.7 N

Explanation:

The density of aluminum is 2710 kg/m³.  So its volume is:

V = (9 kg) / (2710 kg/m³)

V = 0.00332 m³

The apparent weight is the actual weight minus the buoyant force.

N = mg − B

N = mg − ρVg

N = g (m − ρV)

N = (9.8 m/s²) (9 kg − (1000 kg/m³) (0.0332 m³))

N = 55.7 N

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Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?
sammy [17]

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7 0
3 years ago
A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the
Alecsey [184]

Answer:

Part a)

KE = 77.95 J

Part b)

L = 3.16 m

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2

KE = \frac{7}{10} mv^2

KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)

KE = 77.95 J

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

mgLsin\theta = KE

\frac{42}{9.81}(9.81)L sin36 = 77.95

L = 3.16 m

Part c)

by equation of energy conservation we know that

\frac{7}{10}mv^2 = mgL sin\theta

so here we can see that distance L is independent of the mass of the sphere

7 0
3 years ago
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