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Tatiana [17]
3 years ago
10

Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the

snowmobile, including Sully, is 280 kilograms. If Sully accelerates to a velocity of 16 meters/second over 10 seconds, what’s the force exerted by the snowmobile to accelerate? Use F = ma, where . 160 N 168 N 248 N 280 N 324 N
Physics
2 answers:
Gelneren [198K]3 years ago
7 0
Answer: 168N

16 - 10 = 6
6 / 10 = .6
F = 280 x .6 = 168
Svetradugi [14.3K]3 years ago
6 0
16 - 10 = 6
6 / 10 = .6 (this sign / means to divide)
F = 280 x .6 = 168
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8 0
3 years ago
A man walks 30 m to the west, then 5 m to the east in 45 seconds. What is his average speed?
dsp73

The average speed is  0.55 m/s

<h3>What was his displacement?</h3>

Displacement is defined as the change in position of an object.

Mathematically,

D = Initial value - Final Value

D = 30 - 5 = 25m

Now , We have to find the average velocity,

Average velocity is calculated by the formula V = D/t,

where V equals the average velocity, D equals total displacement and t equals total time.

V = D/t

V = 25 m / 45s

V = 0.55 m/s

So , The average speed is o.55 m/s

learn more about average speed:brainly.com/question/9834403

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2 years ago
When a warm air mass moves into an area where a cold air mass is located, it is called a(n) _____?
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8 0
4 years ago
Two children hang by their hands from the same tree branch. the branch is straight, and grows out from the tree trunk at an angl
Ratling [72]

The net torque exerted by the children on the branch of the tree is 1382 N-m.

The torque exert by the kids is calculated as

T= 45.6*9.8*1.28*cos27.5°+36*9.8*(2.25-1.28)cos27.5°

T=1382 N-m

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7 0
4 years ago
Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY &amp; SMALL TIME OF FLIGHT) will
Gennadij [26K]

Answer:

Explanation:

The horizontal distance traveled by the projectile is given by the formula

R=\frac{u^{2}Sin2\theta }{g}

The formula for the time of flight is given by

T = \frac{2u Sin\theta }{g}

Case I: when the launch angle is 30°

So, R_{1}=\frac{u^{2}Sin60 }{g}

R_{1}=\frac{0.866u^{2}}{g}

Horizontal velocity = u Cos 30 = 0.866 u

T_{1} = \frac{2u Sin30 }{g}=\frac{u}{g}

Case II: when the launch angle is 60°

R_{2}=\frac{u^{2}Sin120}{g}

R_{2}=\frac{0.866u^{2}}{g}

Horizontal velocity = u Cos 60 = 0.5 u

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By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2

5 0
3 years ago
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