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Maksim231197 [3]
3 years ago
12

1

Physics
1 answer:
Marina86 [1]3 years ago
4 0

Explanation:

yooooooooo

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A projectile is launched at an angle of 60° from the horizontal and at a velocity of
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A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
soldier1979 [14.2K]

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

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3 0
3 years ago
A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

8 0
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