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3 years ago

Need asap!

Physics

1 answer:

Genrish500 [490]3 years ago

7 0

Answer:

3.0 A Is the correct option

Explanation:

I = V / R

I = 12/ 4

= 3

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A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

$D_l=d$

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

$D_l=d$

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3 years ago

Why are men typically less stable on their feet than women

Viktor [21]

Women generally have a lower centre of gravity than men, contributing to greater stability. Men generally have more muscle mass in their upper bodies,

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3 years ago

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A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in

Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

for the first one: 

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

for the second one :

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

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a=3.33

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He steel used to make structural beams in buildings is a mixture. What category best describes this type of mixture?

aleksklad [387]

The answer would be heterogeneous mixture

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

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3 years ago