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3 years ago

Select the correct answer.

Physics

1 answer:

Rudik [331]3 years ago

7 0

Answer:

B. 3.24 m/s²

Explanation:

We can only answer if we make some assumptions.

ASSUMING

1) The box is on a horizontal surface

2) The surface is frictionless (This will give the MAXIMUM acceleration possible)

F = ma

100cos30 = 25a

a = 100cos36 / 25 = 3.23606797...

a ≈ 3.24 m/s

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A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t

nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$ , Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V

4 0

3 years ago

Read 2 more answers

What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub

aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse ( $F \times t$ ). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

$s=ut+\frac{1}{2} gt^2$

Here, u is initial velocity which is zero.

$s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }$ .

Thus, impulse

$= F \times \sqrt{\frac{2s}{g} }$

From Newton`s second law,

$F =mg$

Therefore, impulse

$= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}$

Given, $m = 7.3 kg$ and $s = 2.0 m$

Substituting these values, we get

Change in momentum = impulse

$= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns$ .

8 0

2 years ago

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

$a=2.5 m/s^2$

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

$d=0+\frac{1}{2}(2.5)(35)^2=1531 m$

8 0

3 years ago

I don’t get science at all my teacher doesn’t explain anything and I was out of school can anyone explain this to me

bekas [8.4K]

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5 0

2 years ago

What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?

Archy [21]

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

$I=\dfrac{E^2}{2\mu_{0}c}$

$E=\sqrt{2\times I\mu_{0}c}$

Put the value into the formula

$E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}$

$E=931.17\ N/C$

Using formula of magnetic field

$B = \dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{931.17}{3\times10^{8}}$

$B=0.0000031039\ T$

$B=3.10\times10^{-6}\ T$

(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using gravitational force

$F=\dfrac{Gm_{s}M_{e}}{r^2}$

Put the value into the formula

$F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}$

$F=3.522\times10^{22}\ N$

We need to calculate the radiation force

Using formula of force

$F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}$

Put the value into the formula

$F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2$

$F_{R}=4.8\times10^{8}\ N$

The gravitational and solar electromagnetic pressure forces of the sun on the earth

$\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}$

$\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}$

Hence, This is the required solution.

3 0

3 years ago