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3 years ago

8

an artificial satellite of mass 900 kg is lunched at a speed of 11,000m/s from its launching station. how much is the kinetic en

ergy imparted to it?

1 answer:

just olya [345]3 years ago

4 0

Answer:

54,450,000,000 Joules

Explanation:

Kinetic Energy = $\frac{1}{2} mv^2$

$\frac{1}{2}$ x $900$ x $(11000)^2$ = 54,450,000,000 Joules

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Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 708-k

Varvara68 [4.7K]

Answer:

a) $I=646.9298\ A$

b) $I=1942.018\ A$

Explanation:

Given:

efficiency of the motor, $\eta=0.95$

voltage of the battery, $V=12\ V$

mass of the car, $m=708\ kg$

a)

initial velocity, $u=0\ m.s^{-1}$

final velocity, $v=25\ m.s^{-1}$

time taken for the acceleration, $t=1\ min=60\ s$

Now we know by the Newton's second law of motion:

$F=m.a$

$F=708\times \frac{(25-0)}{60}$

$F=295\ N$

Now the power will be :

$P=F.v$

$P=295\times 25$

$P=7375\ W$

<u>According to the question:</u>

0.95 times of the electrical power should yield this mechanical power.

$P=V.I\times 0.95$

$7375=12\times I\times 0.95$

$I=\frac{7375}{12\times 0.95}$

$I=646.9298\ A$

b)

height climbed by the car, $h=200\ m$

velocity of climb, $v=25\ m.s^{-1}$

time taken to climb the height, $t=2\ min=120\ s$

force exerted to overcome air and frictional resistances, $f=423\ N$

Now the Power required to climb the hill:

$P=\frac{m.g.h}{t} +f\times v$

$P=\frac{708\times 9.8\times 200}{120}+423\times 25$

$P=22139\ W$

<u>Now according to the electrical efficiency:</u>

$P=V.I\times 0.95$

$22139=12\times I\times 0.95$

$I=1942.018\ A$

7 0

4 years ago

Read 2 more answers

How is amplitude related to volume?

77julia77 [94]

Finally, the amount of increased volume<span> that is achieved by an increase in </span>amplitude<span>is also affected by the frequency. So, turning the </span>amplitude<span> level in the mid-range sound from the previous example will not have the same proportional effect on loudness as the bass sound.</span>

7 0

3 years ago

Read 2 more answers

Help me with this please

Tanya [424]

Answer:

century eh i

Explanation:

8 0

3 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

What average net force is required to accelerate a 9.5 g

Nesterboy [21]

Answer:

2361 Newtons

Explanation:

From the second Newton's law of motion;

F = ma

In this case;

we are given;

Mass as 9.5 g

Initial speed as 0 m/s

Final velocity as 650 m/s

Distance is 0.85 m

Using the equation;

V² = U² + 2as

But u = 0

v² = 2as

Therefore;

a = v² ÷ 2s

= 650² ÷ 2(0.85)

= 248,529.40 m/s²

But;

F = ma

= 0.0095 kg × 248,529.40 m/s²

= 2361 Newtons

Therefore;

The average net force required to accelerate the bullet is 2361 Newtons.

6 0

3 years ago