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m_a_m_a [10]
3 years ago
7

A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resis

tance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter
Physics
1 answer:
miv72 [106K]3 years ago
4 0

Answer:

Explanation:

For resistance , the expression is as follows .

R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .

cross sectional area = π x ( .5 x 10⁻³ )²

S = .785 x 10⁻⁶ m²

Putting the values

R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶

= 1.796 x 10⁻² ohm .

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What is the predicted result when two different materials of equal mass absorb the same amount of energy by heat flow?
Ludmilka [50]

The correct is D.

Explanation: The specific heat is defined as heat required to raise the temperature of a unit mass by one degree. Greater the specific heat, more is the heat required to raise the temperature for equal mass. So, the temperature of the material with lowest specific heat will increase the most for the same amount of heat energy.

7 0
3 years ago
A person travelled 350 m east from his home and returns back home an hour has displacement of_?​
Svetradugi [14.3K]

Answer:

vector of zero magnitude

Explanation:

The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.

In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.

If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude

7 0
2 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106hz. calculat
lakkis [162]
When it comes to wave behavior, there are parameters called wavelength and frequency. These two are related by speed of the radiowave. Radiowaves are electromagnetic waves which travels as fast as light. The wavelength is the distance while frequency is the reciprocal of time. When you multiply them both, you get the electromagnetic wave's speed. The equation is c = wavelength*frequency, where c is the speed of light equal to 3 x 10^8 m/s. 

3 x10^8 m/s = wavelength/104.9 x 10^6 Hz   (Hertz is 1/s)
wavelength = 2.86 meters
6 0
3 years ago
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio
lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

W = F\times d

KE = 0.5\times m\times v^2

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

W = F \times d = 2.62 N \times 100 m

W = 261.6 N\times m

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

0.5 m\times v2^2 = 0.5 m\ v1^2 - W

Now solve for v2

v2 = \sqrt{v1^2 - {\frac{2W}{M}}}

= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Ff = Us\times N

Now solve for Us

= \frac{Ff}{N}

= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}

= 0.00494

4 0
3 years ago
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