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3 years ago

13

An air craft heads. north at 400 km/hrs relative to the wind the air craft velocity is 300 km/hrs due north .find the velocity o

f the wind to the ground.

1 answer:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

325

Explanation

seio na:

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How much force is needed to accelerate a 1,500 kg car at a rate of 7 m/s2?

Sidana [21]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)
   | |
   (1,500 kg) x (7 m/s²) = 10,500 newtons .

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4 years ago

A ball is launched straight up with initial speed of 30.0 m/s. What is the ball's velocity when it comes back to its original po

Zanzabum

Answer:

We could get the time taken by the ball to return back to earth, using the formula:

s = u t + ½ a t², where

s = displacement of the body moving with initial velocity u, acceleration 'a' in time t.

In the present case s=0 (as the ball returns back to starting time)

u= 30 m/s; a = -10 m/s² ( negative sign as a is in opposite direction to u); t=?

0 = 30 t - ½ ×10 ×t²; ==> 5 t = 30, t= 6 second.

So ball will return back after 6 second after being thrown up.

Explanation:

I looked it up

Hope this helps

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3 years ago

A scientist fixes a test charge q′ to point P and then measures the electrostatic force it experiences there. She then calculate

nordsb [41]

Answer:

<em>Correct answer: B.</em>

Explanation:

Electrostatic Field

It measures the electric effect of a charge distribution in its surroundings. If we wanted to test or measure the electric field by using a point-charge of value q', then the electric field is

$\displaystyle E=\frac{F}{q'}$

The electrostatic force between two point charges is

$\displaystyle F=\frac{k\ q_1\ q'}{r^2}$

So, the electric field is

$\displaystyle E=\frac{\frac{k\ q_1\ q'}{r^2}}{q'}$

$\displaystyle E=\frac{k\ q_1}{r^2}$

The theory shows that the electric field doesn't depend on the test charge used, i.e., if we now use q'=-q', the electric force will be of the same magnitude but in the opposite direction, thus the electric field will be the same. Let's recall the formula is used to compute the scalar value of the field, the direction must be studied separately.

Now, if we changed the test charge to another value, say 2q, the measured force will be

$\displaystyle F=\frac{k\ q_1\ 2q}{r^2}$

And the new electric field is

$\displaystyle E=\frac{k\ q_1}{r^2}$

We can see the electric field is not affected by the value of the test charge.

Finally, if we move the test charge to another location and keep the same charge, the electric force will vary and the electric field will be different.

Correct answer: B.

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3 years ago

The __________ wave follows the depolarization wave across the sarcolemma

sammy [17]

Your answer is repolarization

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3 years ago

Distinguish between refraction and diffraction

Anastaziya [24]

Refraction is the change in direction of waves that occurs when waves travel from one medium to another. It is always accompanied by a wavelength and speed of change. Diffraction is the bending of waves around obstacles and openings.

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