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mihalych1998 [28]
3 years ago
12

Most metals are not ?

Physics
1 answer:
mario62 [17]3 years ago
4 0
C liquid at room temperature  
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What is the total resistance in this circuit?
VikaD [51]
Try 40. it seems correct. i’m sorry if i’m wrong.
6 0
2 years ago
How do its eight arms help an octopus obtain food
Andrews [41]
The more arms it has the less of a chance the prey has to swim away.
5 0
2 years ago
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod
cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

F=\dfrac{mv^2}{r}

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0
3 years ago
3. Sailors use a capstan (shown below) to raise and lower the anchor. It takes two sailors located 1 meter from the center to tu
mr_godi [17]

Answer:

B. 2 meters.

Explanation:

To rotate the capstan a certain amount of torque is required, and if each sailor applies a force F at a distance D from the center, then for two sailors the total torque will be

\tau = 2FD\\;

therefore,  for one sailor to apply the same torque it must be that the torque \tau_2 he applies must be equal to the torque that the two sailors applied:

FD_2 =2FD

which gives

D_2 = 2D.

and since D = 1\:meter,

\boxed{D_2 = 2\: meters}

which is choice B.

4 0
3 years ago
A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring project
Gwar [14]

Answer:

x' = 10 x

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

so we will have

v = \sqrt{\frac{k}{m}}(x)

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

5v = \sqrt{\frac{k}{4m}}(x')

now from above two equations we have

\frac{5v}{v} = \frac{x'}{2x}

so we have

x' = 10 x

3 0
2 years ago
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