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shtirl [24]
3 years ago
10

Help with this question please.

Physics
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer: Yes Because it matches with the mass and the amount of force Hope this helps :>

Explanation:

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Is the momentum of a 0.1 kg Mass moving with a velocity of 5 miles per second West
dem82 [27]
<span>Momentum equals Mass x Velocity
Mass equals 0.1kg
Velocity equals 5m/s

So the momentum has to = 0.1 x 5 = 0.5kgm/s

I hope this helped
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Filament bulbs give off lots of __________ that transfers wasted energy to the surroundings. What word completes the sentence?​
Anika [276]

Answer:

filament bulbs give off lots of metal filament that transfers wasted energy to the surrounding.

6 0
2 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 13.1 minutes
ExtremeBDS [4]
Speed and velocity have the same magnitudes. The only difference is that speed is a scalar quantity and velocity is a vector quantity. In other words, speed is just a magnitude, while velocity is a magnitude with direction. They're essentially the same.

Let's convert miles to meters and minutes to seconds

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13.1 minutes = 786 seconds (1 minute = 60 seconds)

Speed is calculated as distance over time, thus,

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b.) Velocity = 4.1 m/s

5 0
3 years ago
The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.
andrew-mc [135]

Answer:

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6 0
2 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0
3 years ago
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