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3 years ago

How are the magnetic interaction & electric-charge interaction similar? Different

1 answer:

3 0

They are similar because they both have opposite poles that attract each other and similar poles that oppose each other. also they are connected because when you run an electric current through some coiled metal it makes a magnet (electro-magnet)

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A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago

An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?

andrey2020 [161]

$From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.$

5 0

3 years ago

The music is in ___meter? a.free time b.duple c.triple d.quadruple

Rashid [163]

D.quadruple is the answer

4 0

3 years ago

Read 2 more answers

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

3 years ago

How does kinetic energy affect the stopping distance of small vehicle compared to a large vehicle

Nadusha1986 [10]

The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.

4 0

4 years ago