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3 years ago

A car moving 20 m/s accelerates at 3 m/s for 5 seconds. what is the final velocity?

Physics

1 answer:

Yuliya22 [10]3 years ago

3 0

The final velocity is of the car is 35 $ms^{-1}$ .

<u>Explanation:</u>

The car initially moves in the velocity 20 $ms^{-1}$ . It then accelerated by 3 $ms^{-2}$ for 5 seconds.

The formula to find final velocity, $v_f=v_i+a(t).$

$v_i$ : initial velocity of the car.

a: acceleration rate.

t: time.

$v_f$ : final velocity

From the given information,

$v_i$ = 20 $ms^{-1}$ .

a=3 $ms^{-2}$ .

t= 5s.

$v_f$ =20+3(5).

=20+15.

=35 $ms^{-1}$ .

∴The final velocity of the moving car is 35 $ms^{-1}$ .

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While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.65 m/s. The stone subs

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Answer:

1. 20.54m/s

2. 1.52s

Explanation:

QUESTION 1:

The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:

v² = u² + 2as

Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

v² = 5.65² + 2 (9.8 × 19.9)

v² = 31.9225 + 2 (195.02)

v² = 31.9225 + 390.04

v² = 421.9625

v = √421.9625

v = 20.5417

v = 20.54m/s

QUESTION 2:

Using v = u + at

Where v = final velocity (m/s) = 20.54m/s

t = time (s)

u = initial velocity (m/s) = 5.65m/s

a = acceleration due to gravity (m/s²)

v = u + at

20.54 = 5.65 + 9.8t

20.54 - 5.65 = 9.8t

14.89 = 9.8t

t = 14.89/9.8

t = 1.519

t = 1.52s

3 0

3 years ago

A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan

ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of $a \;\;m/s^2$ from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, $a=\frac{v-u}{t}=\frac{5-4.2}{t}$

$\Rightarrow at=0.8\cdots(i)$

Now, from the equation of motion

$s=ut+\frac 12 at^2$

$\Rightarrow s=ut+\frac 12 at(t)$

$\Rightarrow 115=4.2t+\frac 12 \times 0.8 t$ [ from equation (i)]

$\Rightarrow 115=(4.2+0.4)t$

$\Rightarrow t= 115/4.6 = 25$ seconds.

Hence, she takes 25 seconds to walk the distance.

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3 years ago

5) In the last part of step 7 of the procedure, you measured the resistance of the flashlight when it had no current passing thr

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Answer:

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