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ValentinkaMS [17]
3 years ago
12

A car accelerates from 21.0 m/s to 46.0 m/s in 2.5 seconds. What is the acceleration of the car.

Physics
1 answer:
vivado [14]3 years ago
6 0

Answer:

10.0 m/s^2

Explanation:

The acceleration of an object is given by

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

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the car in this problem,

u = 21.0 m/s

v = 46.0 m/s

t = 2.5 s

Substituting, we find the acceleration

a=\frac{46.0-21.0}{2.5}=10.0 m/s^2

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The tangent looks good.

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But overall, cool graph.

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3 years ago
1. The process of producing energy by utilizing heat trapped inside the earth's
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Answer:

B

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

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If a wave has a wavelength of 9 meters and a period of 0.006, what is the velocity of the wave?
lana66690 [7]

<u>Answer</u>

D. 1,500 m/s

<u>Explanation</u>

the wave equation states that,

V = λf

Where V ⇒ Velocity

λ ⇒ wavelength

f ⇒ Frequency

F = 1/T

Where T ⇒ period

F = 1/0.006

= 166.667

∴ V = 9 × 166.667

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A model airplane of mass 0.6 kg is attached to a horizontal string and flies in a horizontal circle of radius 6 m, making 1.6 re
nikitadnepr [17]

Answer:

speed= 12.15\frac{m}{s}

Explanation:

In this question we have given

mass of airplane=.6 Kg

Radius of horizontal circle,r=6m

Time taken to complete 1.6 revolution=5s

Therefore time taken to complete 1 revolution, t=\frac{5}{1.6}

t=3.1 s

We have to find the speed of airplane

We will first find the distance covered by airplane in tracing one circle which is equal to circumference of circle

We know that

Circumference,d = 2 \pi \times r

or d=2\times 3.14 \times 6

d=37.68 m

We know that speed=\frac{d}{t}............(1)

Put value of d and t in equation 1

speed=\frac{37.68}{3.1}

speed= 12.15\frac{m}{s}

5 0
3 years ago
A spring with spring constant of 33 N/m is stretched 0.15 m from its equilibrium position. How much work must be done to stretch
Hoochie [10]

Work done is 0.442J

<u>Explanation:</u>

Given:

Spring constant, k = 33 N/m

Distance, x₁ = 0.15m

Additional distance, x₂ = 0.072 m

Total distance = 0.15 + 0.072 m

                        = 0.222 m

Work done, W = ?

We can calculate work done by the formula

W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2

On substituting the value we get:

W = \frac{1}{2}k [(x_2)^2  - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J

Therefore, work done is 0.442J

7 0
3 years ago
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