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laila [671]
3 years ago
6

Which is a possible measurement of the momentum of a moving object?

Physics
2 answers:
Delvig [45]3 years ago
8 0

Momentum is a vector, although we don't hammer on that.  In order to completely describe a momentum, you need a magnitude AND a direction ... just like force and velocity.

So choice #2 is the magnitude of a momentum, without its direction.

<em>Choice #3</em> is  the full package, with both the magnitude and the direction.

Choice #1 has units of energy, and choice #4 has units of acceleration, so neither of those can be it.

VMariaS [17]3 years ago
5 0

choice 1 is a  possible measurement of the momentum of a moving object?


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When does a rubber band, which has been shot at a wall, have the most potential energy?
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D When it is stretched ready to shoot at the wall
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3 years ago
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A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work do
vaieri [72.5K]

Answer:

W= F × d

W= 2kn × 3.6

W= 7.2 J

Work is measured in Joules!

4 0
3 years ago
Someone help please.....
Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean)  = \frac{0.125+0.091+0.0625}{3}

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0
3 years ago
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
3 years ago
I'm willing to give a lot of points if u can help me out.
Anettt [7]

Answer:

i have absolutly no idea how to do it but i looked it up and your answer should be B. i could be wrong but thats what the web told me

7 0
3 years ago
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