3 years ago

DEFINE the term free fall

Physics

2 answers:

bazaltina [42]3 years ago

8 0

Any motion of a body where gravity is the only force acting upon it.

PtichkaEL [24]3 years ago

4 0

Answer:

when you get to fall for free

Explanation:

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Which of these best describes heat? total thermal energy degree of warmth transfer of thermal energy average kinetic energy

Papessa [141]

Total thermal energy is the answer to your question.

3 0

2 years ago

A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?

Irina-Kira [14]

F = 400 Pa x 2.5 m2
F = 1 kN

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2 years ago

Solutions with ions that react with acids or bases to lessen their effects are fubrefs

Softa [21]

Answer:

Buffers

Explanation:

A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.

A buffer solution is an equilibrium system that resists changes in pH or pOH when a small amount of an acid or base is added hence it is a solution of fairly constant pH value.

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3 years ago

A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical

loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0

3 years ago

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For maximum emf, $\sin\omega t=1$

So,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

So, the maximum emf generated in the loop is 5.32 V.

3 0

3 years ago