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3 years ago

Brainliest for the first person to answer Which is a characteristic of an electromagnetic wave?

1 answer:

5 0

The first one. The E and B chatacteristic are perpendicular to eachother. The direction of the wave can be found by the right hand rule.

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evaluate the force of a skateboard exerts on you if your mass is 60kg and you push on the skateboard with a force of 60N

Sav [38]

Every action has an equal or opposite reaction.
You weigh 60kg
<span>So your acceleration is 6N / 60 kg = 0.1m/s^2</span>

6 0

3 years ago

An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti

shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

$I=F\Delta t$

where:

I is the impulse

F is the force exerted on the object

$\Delta t$ is the time during which the force is applied

For the object in this problem, we have

$F=106 N$ (force applied)

$\Delta t= 15.6 s$ (time interval)

Therefore, the impulse experienced by the object is:

$I=(106)(15.6)=1654 kg m/s$

3 0

3 years ago

He likes to eat (into passive voice)

Mama L [17]

Answer:

yes he does

Explanation:

6 0

3 years ago

Find the mass of an object moving at 20 m/s experiencing a momentum of 10<br> kg.m/s

erik [133]

Answer:

Explanation:

The mass of the object can be found by using the formula

$m = \frac{p}{v} \\$

p is the momentum

v is the velocity

From the question we have

$m = \frac{10}{20} = \frac{1}{2} \\$

We have the final answer as

Hope this helps you

3 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