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Lapatulllka [165]
3 years ago
7

Assume that a vaulter is able to carry a vaulting pole while running as fast

Physics
2 answers:
rewona [7]3 years ago
8 0
A,walls
Speleleelelelekeke
Ipatiy [6.2K]3 years ago
8 0

Answer:

5mu = carry a vaulting pole ( 12 m/s) Also

fast

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You are on a ParKour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal
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1.53 meters per second


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3 years ago
A scuba diver and her gear displace a volume of 67.0 l and have a total mass of 64.0 kg. (a) what is the buoyant force on the di
slavikrds [6]
Buoyant force is the force that is a result from the pressure exerted by a fluid on the object. We calculate this value by using the Archimedes principle where it says that the upward buoyant force that is being exerted to a body that is immersed in the fluid is equal to the fluid's weight that the object has displaced. Buoyant force always acts opposing the direction of weight. We calculate as follows:

Fb = W
Fb = mass (acceleration due to gravity)
Fb = 64.0 kg ( 9.81 m/s^2)
Fb = 627.84 kg m/s^2

Therefore, the buoyant force that is exerted on the diver in the sea water would be 627.84 N
4 0
3 years ago
The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The
12345 [234]

Answer:

The electric field is 8.75 \times 10^{6}~v~m^{-1} and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field (\vec{E}) is given by \vec{E} = - \nabla V, 'V' being the potential.

In 1-D, it can be written as

E=\dfrac{V}{d}

where 'd' is the separation of space in between the potential difference is created.

Given, V = 0.070~V~ and the thickness of the cell membrane is d = 8.0 \times 10^{-9}~m.

Therefore the created electric field through the cell membrane is

E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}

5 0
3 years ago
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

4 0
3 years ago
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7. How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a
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The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
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3 years ago
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