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12345 [234]
3 years ago
8

During a collision, a bus runs into a skateboarder inelastically. If the mass of the skateboarders is 70 kg and is moving at 12

m/s to the right, what is the final velocity of the skateboarder if the 2000 kg bus is traveling at 25 m/s to the left before the collision?
Physics
1 answer:
agasfer [191]3 years ago
5 0

The final velocity of the skateboarder is 23.75 m/s.

We can use conservation of momentum to calculate the magnitude of the final velocity.

Momentum before collision is equal to the momentum after collision

m1v1+m2v2=(m1+m2)v

here, v is final velocity of the skateboarder, assuming right side velocity as positive, and left side velocity as negative.

70*12+2000*(-25)=(70+2000)v

v=-23.75 m/s

Here, negative sign indicates that both skyboarder and bus will move in left direction.

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Which nucleus completes the following equation?<br> A. 299 Np<br> B. 20Pa<br> C. 2 Pa<br> D. - Np
blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0
3 years ago
What conclusions can you draw about the connection between the Sun’s altitude and the latitude of the observer on seasonal chang
Cerrena [4.2K]

Answer:

Explanation:

Altitude of the Sun and the latitude position on the earth play an important role in the season change on the earth.

When the altitude of the sun is high then the average temperature of the earth is higher because the luminous intensity of the sun rays is higher due to the focusing of high energy sun rays over a small area.

But when the sun is at higher altitudes we receive less denser rays of the sun and hence we have less heat on the earth on an average.

  • But despite of the altitude some places on the earth have distinct temperature than the other place at the same time of the year. This is due to their latitudinal location. The places near the equator are warmer most of the times throughout the year because they receive the most direct rays while the poles receive slanting rays and hence are colder even in summer when the earth is at lower altitudes.
6 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
True or False:
mart [117]
This is a true statement
3 0
3 years ago
Read 2 more answers
A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m
Alex73 [517]

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

6 0
3 years ago
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