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MakcuM [25]
3 years ago
15

What was the initial speed of a car if its speed is 40 m/s after 5 seconds of

Physics
2 answers:
tester [92]3 years ago
4 0

Answer:

\boxed {\boxed {\sf C. \ 20 \ m/s }}

Explanation:

We are asked to find the initial speed of a car.

We are given the final speed, the time, and the acceleration, so we will use the following kinematic equation:

v_f=v_i+at

We know the final speed is 40 meters per second, the acceleration is 4 meters per second squared, and the time is 5 seconds.

  • v_f= 40 m/s
  • t= 5 s
  • a= 4 m/s²

Substitute the values into the formula.

40 \ m/s = v_i+(4 \ m/s^2 * 5 \ s)

Multiply inside the parentheses.

40 \ m/s =v_i+ 20 \ m/s

We are solving for the initial speed, so we must isolate the variable v_i.  

20 meters per second is being added to v_i. The inverse operation of addition is subtraction. Subtract 20 m/s from both sides of the equation.

40 \ m/s - 20 \ m/s = v_i + 20 \ m/s - 20 \ m/s

40 \ m/s - 20 m/ s = v_i

20 \ m/s=v_i

The initial speed of the car is <u>20 meters per second</u> and <u>choice C</u> is correct.

Monica [59]3 years ago
3 0
We use the formula:
V= u + at

V- final velocity. U- intial velocity. A- acceleration. T- time

Note: velocity is speed in this question

Fill in values:
V= u + at
40= u + 4 x 5
40= u + 20
40 - 20 = u
u = 20m/s
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Jasper made a list of the properties of electromagnetic waves. Identify the mistake in the list. Electromagnetic Wave Properties
Nata [24]

Answer:

Statement 2 is wrong

Explanation:

To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics

* they are transverse waves

* formed by the oscillations of the electric and magnetic fields

* the speed of the wave is the speed of light

with these concepts let's review the final statements

1) True. Formed by the oscillation of the two fields

2) False. They are transverse waves

3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields

4) True. They all have the same speed of light

Statement 2 is wrong

6 0
3 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0
3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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