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borishaifa [10]
3 years ago
13

Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

y of 24 m/s?
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

So, the final velocity is

v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

\Rightarrow v=-15.24 m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

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How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R
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Answer:

This relationship is explained by Ohm's law

Explanation:

Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as

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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

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\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
Which of the following is a contact force?
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The contact force is indeed caused by "Contact". 

Air resistance is basically a type of friction, which is apparently present providing two object contact.

The rest selections are all interaction force, which is not necessarily caused by contact.
4 0
3 years ago
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Divide
       (the distance covered in some period of time)
by
       (the time taken to cover the distance).

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3 years ago
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