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Verizon [17]
3 years ago
10

A certain laboratory experiment requires an aluminum wire of length of 30.0 m and a resistance of 3.80 ω at 20.0°c. what diamete

r wire must be used? (the resistivity of aluminum at 20.0°c is 2.82 ✕ 10−8 ω · m.)
Physics
1 answer:
Tom [10]3 years ago
5 0
Answer:
diameter = 5.316 * 10^-4 meters

Explanation:
The resistance of the wire can be calculated using the following rule:
R = ΡL / A
where:
R is the resistance = 3.8 ohm
Ρ is the resistivity = <span>2.82 * 10^−8 ω · m
</span>L is the length of the wire = 30 m
A is the area that we want to calculate
Substitute with the givens in the above equation to get the area as follows:
3.8 = (2.82 * 10^−8 * 30) / A
A = 2.22 * 10^-7 meters^2
Now, the area of the circle is calculated as follows:
Area = pi * (radius)^2
2.22 * 10^-7 = pi * (radius)^2
(radius)^2 = 7.066 * 10^-8
radius = 2.658 * 10^-4 meters
Diameter is double the radius. This means that:
diameter = 2 * 2.658 * 10^-4 = 5.316 * 10^-4 meters

Hope this helps :)
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Answer:

20m/second

Explanation:

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A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?
yanalaym [24]

Answer:

KE=12,500J

Explanation:

The formula for kinetic energy is:

KE = \frac{1}{2}mv^2

We can plug in the given values into the equation:

KE = \frac{1}{2}*1000kg*(5m/s)^2

KE = 500kg*25m^2/s^2

KE=12,500J

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3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

4 0
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Answer:

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Explanation:

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F = qvB

here we have

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F = 4.8 \times 10^{-13} N

q = 1.6 \times 10^{-19} C

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now from above equation we have

B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}

B = 0.75 T

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