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Len [333]
3 years ago
10

To travel at a constant speed,a car engine provides 24 KW of useful powers. The driving force on the car is 600N.AT what speed d

oes it travel? A 2.5 B 4.0 C.25 D 40
Physics
1 answer:
Whitepunk [10]3 years ago
3 0

Answer: D. 40

Explanation:

Power = Force * Velocity

Force = 600N

Power needs to be converted to Watts:

= 24 * 1,000

= 24,000 W

24,000 = 600 * Velocity

Velocity = 24,000/600

= 40 m/s

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A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice
ladessa [460]

Answer:

Approximately \rm 19.8\; m\cdot s^{-1} (downwards.)

Assumptions:

  • the rocket started from rest;
  • the gravitational acceleration is constantly \rm -9.8\; m \cdot s^{-2};
  • there's no air resistance on the rocket and the two fragments.
  • Both fragments traveled without horizontal velocity.

Explanation:

The upward speed of the rocket increases by \rm 10\; m \cdot s^{-1}. If the rocket started from rest, the vertical speed of the rocket should be equal to \rm 20\; m \cdot s^{-1}.

The mass of the rocket (before it exploded) is 1500 kilograms. At 20 m/s, its momentum will be equal to \rm 20 \times 1500 = 30,000\; kg \cdot m\cdot s^{-1}.

What's the initial upward velocity, u, of the lighter fragment?

The upward velocity of the lighter fragment is equal to v = 0 once it reached its maximum height of x = \rm 530\; m.

v^2 - u^2 = 2g \cdot x.

\begin{aligned}u &= \sqrt{v^2 - 2g\cdot x} \\ &= \sqrt{-2 (-9.8) \times 530}\\ &\approx \rm 101.922\; m \cdot s^{-1}\end{aligned}.

Mass of the two fragments:

  • Lighter fragments: \displaystyle \frac{1}{1 + 2} \times 1500 =\rm 500\; kg.
  • Heavier fragment: \displaystyle \frac{2}{1 + 2} \times 1500 =\rm 1000\; kg.

Initial momentum of the lighter fragment:

m \cdot v = \rm 10192.2\; kg \cdot m \cdot s^{-1}.

If there's no air resistance, momentum shall conserve. The momentum of the lighter fragment, plus that of the heavier fragment, should be equal to that of the rocket before it exploded.

The initial momentum of the heavier fragment should thus be equal to the momentum of the two pieces, combined, minus the initial momentum of the lighter fragment.

\rm 30000 - 10192.2 = 19807.8\;kg \cdot m \cdot s^{-1}.

Velocity of the heavier fragment:

\displaystyle \rm \frac{19807.8\;kg \cdot m \cdot s^{-1}}{1000\; kg} \approx 19.8\; m \cdot s^{-1}.

5 0
4 years ago
Read 2 more answers
Please urgent science question explaining needed
sweet-ann [11.9K]

Answer:

what is the question? there isn't one

3 0
3 years ago
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Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic
dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

F=ILB sin \theta (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, \theta the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"

is correct, because when the current is perpendicular to the magnetic field, \theta=90^{\circ}, sin \theta = 1 and the force is maximum.

Moreover, according to (2), we also see that the statements

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>

because F (the force) is perpendicular to both the magnetic field and the current.

5 0
3 years ago
Fre.e points because you deserve it
bagirrra123 [75]

Answer:

thank you so much.. i appreciate your kindness

5 0
3 years ago
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The reason a charged balloon will stick to a wall is that?
kari74 [83]
The negative charges of the balloon will stick to the positive charges on the wall.
6 0
3 years ago
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