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3 years ago

What is electromagnetic indiction

1 answer:

6 0

Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field. Michael Faraday is generally credited with the discovery of induction in 1831, and James Clerk Maxwell mathematically described it as Faraday's law of induction.

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2. Find the time taken by the bus to reach the stop. need only group B, 2 answer

Molodets [167]

Answer:

t = 2 seconds

Explanation:

In 2nd question, the question is given the attached figure.

Initial speed of the bus, u = 0

Acceleration of the bus, a = 8 m/s²

Final speed, v = 16 m/s

We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :

$a=\dfrac{v-u}{t}$

t is time taken

$t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s$

The bus will take 2 seconds to reach the stop.

3 0

3 years ago

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

In a horse cart system when would there be the smallest amount of acceleration

Assoli18 [71]

Answer:

well... when the horse stops/rests, or if it is blocked by a surface or anything of solid background.

Explanation:

If it is going up a hill or slope and it just starts to move that would also be considered the smallest amount of acceleration this can go for many things when it just starts to move. but I would go for when it rests amounting to your fitting of the question.

3 0

3 years ago

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

vladimir2022 [97]

As we know that friction force on box is given by

$F_s = \mu_s N$