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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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intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

here we know that

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also we know that

r_1 = 10 Ly

now we will have

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

36 = \frac{r_2^2}{10^2}

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8 0
3 years ago
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
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a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

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v(2) = 22 m/s

7 0
3 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

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θ = 30.0°

minimum static friction = ?

now,

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N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

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now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

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7 0
3 years ago
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di
S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F\alphae

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divide all through by extension

1) Total force is

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Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

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1/keq=1/k1+1/k2

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