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rusak2 [61]
3 years ago
10

A 1200-kilogram automobile in motion strikes a 0.0001-kilogram insect. As a result, the insect is accelerated at a rate of 100 m

eters per second squared. What is the magnitude of the force the insect exerts on the car?
Physics
2 answers:
nadya68 [22]3 years ago
6 0

Answer:

force = 1 × 10^{-2} N

Explanation:

given data

automobile mass = 1200 kg

insect mass = 0.0001 kg

insect accelerated = 100 m/s²

to find out

magnitude of the force the insect exerts on the car

solution

we get here force the insect exerts that is express as

force = mass × acceleration    ............1

put here value we get

force = 0.0001 × 100 m/s²

force = 1 × 10^{-2} N

Scrat [10]3 years ago
6 0

Answer:

0.01 N

Explanation:

mass of automobile, M = 1200 kg

mass of insect, m = 0.0001 kg

acceleration, a = 100 m/s^2

Force on insect, F = mass of insect x acceleration

F = 0.0001 x 100

F = 0.01 N

Thus, the force on the insect is 0.01 N.

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2 years ago
A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.
marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

\omega = 25 rpm

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

\omega = 2 \pi f (1)

where

\omega is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

\omega=0.42 rev/s

Keeping in mind that 1 rev = 2\pi rad, the angular speed can be rewritten as

\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42

And re-arranginf eq.(1), we can find the frequency:

f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

v=\omega r

where

\omega is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

\omega = 2\pi \cdot 0.42 rad/s is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

d=vt

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

d=(26.4)(20)=528 cm

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
4 years ago
Read 2 more answers
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