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3 years ago

How long does creatine phosphate provide every

1 answer:

6 0

Creatine phosphate can supply the energy needs of a working muscle at a very high rate, but only for about 8–10 seconds.

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There are several events listed below some of the events are related to tectonic plates activity mark each that is associated wi

iVinArrow [24]

A because an earthquake is shaking of tectonic plates

5 0

3 years ago

A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 0

3 years ago

Please help me please

shusha [124]

Answer:

Explanation:

its A because your comparing so comparative

8 0

3 years ago

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5. You require a high-current battery to start a large tractor. While shopping

Yakvenalex [24]

Answer:

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3 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0

3 years ago