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2 years ago

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An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati

onal potential energy of this object at this position?

1 answer:

IgorC [24]2 years ago

6 0

2400joules

Explanation:

P.E

m= 20kg h=8m g=10m

P.E= 20×8×10

=2400joules

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In a transformer, energy is carried from the primary coil to the secondary coil by:________

likoan [24]

In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>

An item utilized in the transfer of electric energy is a transformer.
AC current is used for transmission.
It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
EMF is therefore generated in the secondary coil.

Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

Learn more about the transformer here:

brainly.com/question/26787198

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2 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

3 0

3 years ago

You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

4 0

3 years ago

A 1,650 kg SUV comes uniformly to a stop. If the vehicle is accelerating at -1.3 m/s^2,

lions [1.4K]

Answer:b

Explanation:

4 0

3 years ago

A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in

Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0

3 years ago