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2 years ago

7

A mosquito flying over a highway strikes a windshield of a moving truck. Compared to the magnitude of the force of the truck on

the mosquito during collision, the magnitude of the force of the mosquito on the truck is
Larger
Smaller
The Same
Cant be determined

1 answer:

krok68 [10]2 years ago

3 0

Answer:

the answer to this question is

<em>The</em><em> </em><em>Same</em><em> </em>

<em>newton's</em><em> </em><em>law</em><em> </em><em>#3</em>

Explanation:

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em>

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Tyler has learned that potential energy is energy stored. Tyler's teacher asks the students to come up with a demonstration of p

Advocard [28]

D.A child at the top of a slide

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2 years ago

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Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0

2 years ago

An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?

postnew [5]

To finish one orbit it will take 98 x 60 seconds. So; <span>(2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec. </span><span>
</span>

8 0

2 years ago

A marble is dropped straight down from a distance h above the floor.

Mrac [35]

Fm=Fe and am>ae
Hopefully this helps

7 0

2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

2 years ago