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Delvig [45]
3 years ago
8

Make the following conversion

Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

The value of 60.2 mg in to g is 0.0602.

Explanation:

We need to calculate the 60.2 mg in to gram.

Conversion mg to g :

1 g = 1000 mg

Using this conversion,

1 mg = 10^{-3}g

So, 60.2\ mg = 60.2\times10^{-3}\ g

60.2\times10^{-3}\ g

Hence, The value of 60.2 mg in to g is 0.0602.

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Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
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Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

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F_A-\mu mg=ma

\dfrac{F_A-\mu mg}{m}=a

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a=-3.905\ m/s^2

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A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
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Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

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