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LenaWriter [7]
3 years ago
7

How can voltage be induced in a wire with the help of a magnet?

Physics
1 answer:
babunello [35]3 years ago
3 0
If you move a magnet through a loop of wire, induction will happen. The more loops you make, the stronger the effect becomes.
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juan aims his boat directly across a river flowing at 8mi/hr. juan's boat travels at 6 miles/hr in still water. how was does jua
son4ous [18]
6mph:

Suppose the boat is traveling on a y axis. The river flow acts on the x axis. Motion on each axes are independent. The linear speed of the boat is not changed. Furthermore the projectile motion is changing, but you're specifically asking about the linear speed of the boat which is unchanged.
4 0
3 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
PLS ANSWER FAST WILL MARK BRAINLIEST
densk [106]

In position A

<em>The</em><em> </em><em>mirror </em><em>has</em><em> </em><em>to</em><em> </em><em>be</em><em> </em><em>observed</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>ray</em><em> </em><em>reflected</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>mi</em><em>rror</em><em>.</em>

<em>Since</em><em> </em><em>the</em><em> </em><em>the</em><em> </em><em>angle</em><em> </em><em>of</em><em> </em><em>reflec</em><em>tion</em><em> </em><em>is</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>angle</em><em> </em><em>of</em><em> </em><em>incidence</em><em>,</em><em> </em><em>the</em><em> </em><em>reflected</em><em> </em><em>ray</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>in</em><em> </em><em>equal</em><em> </em><em>path</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>incident</em><em> </em><em>ray</em><em>.</em>

<em>That</em><em> </em><em>phenomenon</em><em> </em><em>is</em><em> </em><em>only</em><em> </em><em>in</em><em> </em><em>diagram</em><em> </em><em>A</em>

7 0
3 years ago
A beam of yellow light is made to pass through two slits that are 3.0 x 10−3 meters apart. On a screen 2.0 meters away from the
lakkis [162]

Answer:

585 nm

Explanation:

The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:

y=\frac{m\lambda D}{d}\Delta y =\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength

D is the distance of the screen from the slits

d is the separation between the slits

The distance between two consecutive bright fringes therefore is given by:

\Delta y = \frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}

In this problem we have:

\Delta y = 3.9\cdot 10^{-4} m (distance between two bright fringes)

D = 2.0 m (distance of the screen)

d = 3.0 x 10−3 m (separation between the slits)

Solving for \lambda, we find the wavelength:

\lambda=\frac{\Delta y d}{D}=\frac{(3.9\cdot 10^{-4})(3.0\cdot 10^{-3})}{2.0}=5.85\cdot 10^{-7} m = 585 nm

4 0
3 years ago
How to charge a laptop battery without a charger?
Likurg_2 [28]
I dont know if it right but <span>Use a universal power adapter. This is perhaps the most obvious solution to your battery woes. Readily found at most retail outlets that carry electronics, a universal power adapter can range anywhere in price from $30 to $100 or more. The adapter comes with multiple tips, one of which will likely fit your laptop’s charging port. When plugged in, the adapter will not only power your laptop, but will charge its battery as well.</span>
7 0
3 years ago
Read 2 more answers
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