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grandymaker [24]
3 years ago
12

A steel wire of cross_ sectional area 2x10-5 m2 is stretched through 2mm by a force of 4000 N. find the young's modulus of the w

ire . the length of the wire is 2m.
please answer me.​
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer:

hi how are you doing today Jasmine

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A tungsten wire has resistance R at 20°C. A second tungsten wire at 20°C has twice the length and half the cross-sectional area
Bad White [126]
The resistance is 4 times the resistance of the first wire. the formula is R = p*l/A with p being resistivity, l length and A area. So if you double length and half area, which botv result in more resistance, you get p*2/0.5 or 4 (p can be abandoned because it is the same. We take standard length and area as 1)
6 0
3 years ago
Suppose you wish to construct a motor that produces a maximum torque whose magnitude is 1.7 × 10-2 N·m. The coil of the motor ha
Len [333]

Answer:

86 turns

Explanation:

Parameters given:

Magnetic torque, τ = 1.7 * 10^(-2) Nm

Area of coil, A = 9 * 10^(-4) m²

Current in coil, I = 1.1 A

Magnetic field, B = 0.2 T

The magnetic toque is given mathematically as:

τ = N * I * A * B

Where N = number of turns

To find the number of turns, we make N subject of formula:

N = τ/(I * A * B)

Therefore:

N = (1.7 * 10^(-2)) / (1.1 * 9 * 10^(-4) * 0.2)

N = 85.85 = 86 turns (whole number)

The number of turns must be 86.

3 0
4 years ago
Shari uses the data in the table to predict how her weight on Venus would compare with her weight on Earth.
spin [16.1K]

Answer :  <em>Her weight is lower on Venus because the acceleration due to gravity is lower.</em>

Explanation :

Venus is also called as Earth's twin. This is because both the mass and the size of Earth and Venus are almost same. The acceleration due to gravity on earth is 9.8\ m/s^2 while on Venus is 8.87\ m/s^2.

So, when Shari measure her weight on Venus she found her weight is lower on Venus. This is because the acceleration due to gravity is lower on the surface of Venus as compared to the Earth.

Since,    w=mg

i.e. weight depends on g.

<em>So, correct prediction is (b)</em>

3 0
3 years ago
Read 2 more answers
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
4 years ago
An object is five focal lengths from a concave mirror.how do the object and image heights compare?
enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

7 0
3 years ago
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