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Agata [3.3K]
2 years ago
7

A car increases its speed from 20km/hr to 50km/hr in seconds. Its acceleration is _____.

Physics
1 answer:
mash [69]2 years ago
7 0

Answer:

Option D

0.83 m/s2

Explanation:

Time is assumed as 10 seconds

First, convert the speeds from km/h to m/s

20 km/h\times \frac {1000 m}{3600 s}=5.555555556 m/s \approx 5.56 m/s

50 km/h\times \frac {1000 m}{3600 s}=13.88888889  m/s \approx 13.89 m/s

Acceleration, a=\frac {v-u}{t} where u and v are the initial and final velocities respectively, t is the time taken to accelerate.

Substituting 13.89 m/s for v, 5.56 m/s for u and 10 s for t then

a=\frac {13.89-5.56}{10}=0.8333333 m/s^{2}\approx 0.83 m/s^{2}

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A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
rusak2 [61]

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

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8 0
2 years ago
There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat
Mamont248 [21]

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator =heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec

(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

Let the temperature be \Delta T

Heat absorbed per hour =504000 [tex]=400\times 10^3\times \Delta T

So  \Delta T=\frac{504000}{400000}=1.26K

8 0
3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
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The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
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Let's use the other equation to find the time
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              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
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