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Paladinen [302]
3 years ago
13

A 300 kg car travels at 11 ms. What is the momentum

Physics
1 answer:
sweet-ann [11.9K]3 years ago
5 0
The momentum of a <span>300 kg car traveling at 11 ms is 3300 ms</span>
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A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?
Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

4 0
3 years ago
An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?​
snow_lady [41]
The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

7 0
2 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
Can any kind soul help me please​
Alex

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

3 0
3 years ago
A person makes a cup of coffee by first placing a 200 W electric immersion heater in 0.32 kg
alexdok [17]

Answer:

Answer is D. 8.04 x 10^4 J

Explanation:

1. D

2. A

3. D

4. B

5. C

6. B

7. D

8. C

9. B

10. D

All correct i promise you that

8 0
3 years ago
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