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3 years ago

Help with this please

Physics

1 answer:

kompoz [17]3 years ago

3 0

An airplane requires a much less force in order to get off the ground than a rocket does, since a rocket needs to exit Earth’s atmosphere.

Planes typically travel at much slower speeds that traditional rockets, as they are faced with varying restrictions that affect how quickly they fly.

Both forms of transportation, although very different in certain aspects, are similar in the fact that they both need to adhere to similar constraints. Although most airplanes aren’t faced with the challenge of exiting the atmosphere, they do need to focus on their fuel, their safety, etc. just as rockets do as well.

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Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the

den301095 [7]

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

4 0

3 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

4 0

2 years ago

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

$F=ma$

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0

2 years ago

Read 2 more answers

What was your train of thought as you navigated the picture of the candle?

defon

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

4 0

3 years ago

On the kelvin scale what is the freezing point of water

Alla [95]

Freezing point of the water is known as 273 K

Hope this helps!

8 0

3 years ago