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Verizon [17]
3 years ago
5

Why do we use the name electromagnetism to describe the force of an

Physics
1 answer:
ElenaW [278]3 years ago
5 0

Answer:

magnet, gravity

Explanation:

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Which best describes the forces identified by Newton’s third law of motion?
egoroff_w [7]
<span>equal and acting on different objects</span>
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Please help on this one?
butalik [34]

B. +Q, + W are the correct sign

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A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t
borishaifa [10]

Answer:

Angular acceleration will be 18.84rad/sec^2

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity \omega _i=0rev/sec

And final angular velocity \omega _f=24rev/sec

Time is given as t = 8 sec

From equation of motion

We know that \omega _f=\omega _i+\alpha t

24=0+\alpha \times 8

\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2

So angular acceleration will be 18.84rad/sec^2

4 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
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