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sp2606 [1]
3 years ago
12

A large truck is moving at 22.0 m/s. If its momentum is 125,000 kg • , what is the truck's mass?

Physics
2 answers:
PIT_PIT [208]3 years ago
6 0
The mass of the truck is 5680 kg
alexandr402 [8]3 years ago
6 0

Answer:

5682 kg

Explanation:

The momentum of a moving object is given by:

p=mv

where m is the mass of the object while v is its velocity.

In this problem, we know the velocity of the truck (v=22.0 m/s) and its momentum (p=125,000 kg m/s), therefore we can rearrange the previous formula to calculate the mass of the truck:

m=\frac{p}{v}=\frac{125,000 kg m/s}{22.0 m/s}=5682 kg

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What is a partial and total lunar eclipse?
Minchanka [31]

Partial Lunar Eclipse:

A partial lunar eclipse is when the earth gets between the Sun and Moon. However, all three bodies are not in alignment meaning we are able to see some more like part of the moon's surface as it moves in route of the Earth's shadow.

Total Lunar Eclipse:

The three celestial bodies are perfectly aligned which allows for the earth to completely block the sun's rays from hitting/reaching the moon. The sun is positions is in back of the Earth which then causes the shadow of the earth to be cast on the Moon covering the moon completely. When that happens you get the phenomenon called a total lunar eclipse.

Hopefully this helped and good luck.



7 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2
MArishka [77]
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
               =2.905 m/s
Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
              = 0.296 seconds


3 0
3 years ago
Which factors could be potential sources of error in the experiment? check all that apply.
Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

6 0
3 years ago
Read 2 more answers
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

5 0
3 years ago
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