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3 years ago

What is an interaction pair?

Physics

1 answer:

Alchen [17]3 years ago

6 0

A set of two forces that are in opposite directions, have equal magnitudes and act on different objects

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

v is the speed

r is the radius

Now,

$a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2$

In g units:

$a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g$

7 0

3 years ago

Coders play an important role in

Burka [1]

Answer:

it is a.health record documentation

Explanation:hope this helps

3 0

3 years ago

Which conclusion is supported by the data in the table?

vaieri [72.5K]

Answer:

B. More African Americans became registered to vote in Southern states.

Explanation:

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3 years ago

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4

n200080 [17]

Answer:

Explanation:

Let v is the launch speed of the plastic ball and the angle of projection is θ.

So, in horizontal direction

v Cosθ x t = 4.8 .... (1)

In th evertical direction

1.4 = v Sin θ x t - 0.5 gt² .... (2)

As , v Sin θ x t = 3.8 .... (3) , put in equation (2)

1.4 = 3.8 - 4.9 t²

t = 0.7 s

Put in (1) and (3)

v Cosθ x 0.7 = 4.8

v Cosθ = 6.86

and v Sinθ x 0.7 = 3.8

v Sinθ = 5.43

Now

$v^{2}\left ( Sin^{2}\theta +Cos^{2}\theta \right )=5.43^{2}+6.86^{2}$

v = 8.75 m/s

3 0

3 years ago

Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di

Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, $u=20m/s$

And the mass of the car is, $m=1000 kg$

The total distance covered by the car before stop, $s=45m$

And the final speed of the car is, $u=0m/s$

Now initial kinetic energy is,

$KE_{i}=\frac{1}{2}mu^{2}$

Substitute the value of u and m in the above equation, we get

$KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J$

Now final kinetic energy is,

$KE_{f}=\frac{1}{2}mv^{2}$

Substitute the value of v and m in the above equation, we get

$KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J$

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

$F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN$

Here, the force is negative because the force and acceleration in the opposite direction.

6 0

3 years ago