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kobusy [5.1K]
3 years ago
15

In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

ld. The emf induced in the coil is 5.8 V. Other things being equal, what would the induced emf be if the field had a magnitude of 0.55 T?
Physics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}

\frac{5.8}{e_{2}}=\frac{0.38}{0.55}

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

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Kyle is wondering what he is getting for his birthday. He decides to shake the box that contains his present. Kyle did not reali
stiks02 [169]

Answer: Asking questions

Explanation:

Scientists always start with their question before

Observing anything

8 0
3 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
A 2.6 kg mass attached to a light string rotates on a horizontal,
Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

  • <em>maximum mass the string can support before breaking, m = 17.9 kg</em>
  • <em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0
2 years ago
On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

          =\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119

          = 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200
SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

Dd

>

1 and your displacement from the store to your friend’s house

be Dd

>

2.

Given: Dd

>

1 = 200 m [N]; Dd

>

2 = 600 m [S]

Required: Dd

>

T

Analysis: Dd

>

T 5 Dd

>

1 1 Dd

>

2

Solution: Figure 6 shows the given vectors, with the tip of Dd

>

1

joined to the tail of Dd

>

2. The resultant vector Dd

>

T is drawn in red,

from the tail of Dd

>

1 to the tip of Dd

>

2. The direction of Dd

>

T is [S].

Dd

>

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

>

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0
2 years ago
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