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Liula [17]
3 years ago
9

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

of this cheetah? a. 6.5 m/s^2 b. 7.2 m/s^2 c. 5.6 m/s^2 d. 5.8 m/s^2 e. 7.7 m/s^2
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration,a=\frac{v-u}{t}

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

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Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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3 years ago
A camera employs a ............................... lens to form...................... images. a. converging .... real b. converg
olga2289 [7]

Answer:

b

Explanation:

4 0
3 years ago
What force is required to give an object with mass 125 kg an acceleration of 3
stiks02 [169]
The answer is B

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4 years ago
7) You think you have found a diamond. Its mass is 5.28 g and its volume is 2 cm3. Calculate the density
lesya [120]

Answer:

\boxed {\tt d=2.64 \ g/cm^3}

\boxed {\tt Not \ a \ diamond}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass is 5.28 grams and the volume is 2 cubic centimeters.

m=5.28 \ g\\v= 2 \ cm^3

Substitute the values into the formula.

d=\frac{5.28 \ g }{2 \ cm^3}

Divide.

d=2.64 \ g/c^3

The density of the unknown substance is 2.64 grams per cubic centimeter.

The density of a diamond is about 3.5 grams per cubic centimeter. Since 2.64 is not equal to 3.5, the unknown substance is not a diamond.

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I can see three different transitions here:

3 --> 1

3 --> 2
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