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ra1l [238]
4 years ago
13

Which value would complete the last cell?

Physics
1 answer:
GREYUIT [131]4 years ago
5 0

Answer:

4.0

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as

Force (F) = mass (m) x acceleration (a)

F = ma

With the above formula, we can obtain th acceleration of the body as follow:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

F = ma

20 = 5 x a

Divide both side by 5

a = 20/5

a = 4 m/s²

Therefore, the value that will complete the last cell in the question above is 4.

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An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai
rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0
3 years ago
g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for
lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

a\text{sin}\theta=\frac{2N+1}{2}\lambda

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}

Solving the above equation we get

we N= 2.313≅ 2

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3 years ago
Which equation can be used to calculate the normal force on an object if you know the speed of the object, the coefficient of ki
tino4ka555 [31]

Answer:

D

Explanation:

The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.

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2 years ago
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