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sergiy2304 [10]
3 years ago
8

a toy car travels from A to B at constant speed 30km/hr and without stopping at B return A at constant speed v. if the average s

peed of the car is 24km/h then v is​
Physics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

Speed BA = 18 km/hr.

Explanation:

Given the following data;

Speed AB = 30km/hr

Speed BA = x km/hr

Average speed = 24 km/hr

To find the value of x;

Average speed = (Speed AB + Speed BA)/2

Substituting into the equation, we have

24 = (30 + x)/2

48 = 30 + x

x = 48 - 30

x = 18 km/hr

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The study of charges in motion and their
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What is an observation in an experiment ?
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3 years ago
How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m
Mars2501 [29]

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2

Net force, F = ma

F=1600\times 1.56\\\\F=2496\ N

So, the net force is 2496 N.

6 0
3 years ago
Usually wire resistance can be neglected. But we see the effect of it sometimes when a light dims as another high-current device
tangare [24]

Please find the figure attached below:

Answer:

a.Resistance of bulb=R_{2}=192 ohm

b.power consumed by bulb after motor disconnection=74.609 W

Explanation:

<u>a.Resistance of bulb=</u>R_{2}<u>=?</u>

As we know that P=\frac{P^{2} }{R}

putting R as resistance of bulb i.e. R=R_{2}

P=\frac{V^{2}}{R_{2} }\\R_{2}=\frac{V^{2}}{P} \\ R_{2}=\frac{(120)^{2} }{75}\\ R_{2}=192ohm

<u>b.power consumed by bulb after motor disconnection? </u>

from the figure we see that R_{1}\ and\ R_{2} are in series so

R_{eq}=R_{1} +R_{2} \\R_{eq}=192+0.5\\R_{eq}=192.5ohm

current through their resistance is

I_{eq}=\frac{V}{R_{eq} }\\ I_{eq}=\frac{120}{192.5 } \\I_{eq}=0.6233A

Power consumed by bulb is

P_{b} =I^{2}R\\\\ P_{b} =(0.6233)^{2}(192)\\\\P_{b} =74.609W

3 0
4 years ago
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