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Bezzdna [24]
2 years ago
7

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

iginally at 24°C, neglecting any changes in dimensions? Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is originally at 24°C, neglecting any changes in dimensions?
Physics
1 answer:
sleet_krkn [62]2 years ago
8 0

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

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crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

<u>f = 485.62 N</u>

6 0
3 years ago
A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .
sp2606 [1]
How much work in J does the string do on the boy if the boy stands still? 

<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>

<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>

<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>

<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>

<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
6 0
3 years ago
7. What is the velocity of an object with a distance of 90m south and a time of<br> 5s?
IrinaK [193]

Answer:

Explanation:

v= s/t

V =90m/5s

V = 8m/s

4 0
2 years ago
A car starts from rest and after 20 seconds it's velocity becomes 108km find the acceleration of the car
andre [41]

Answer:

1.5 km/s²

Explanation:

Given that:

a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s

after time (t) = 20 seconds

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The acceleration (a) of the car can be determined by using the formula:

a = \dfrac{v-u}{t}

a = \dfrac{30\  m/s -0 \ m/s}{20 \ s}

a = \dfrac{30 \  m/s}{20 \ s}

a = 1.5 km/s²

7 0
3 years ago
Suppose you lift a 20 kg box by a height of 1.0 m.
9966 [12]
W=mgh W=(20)(9.8)(1) w=196J
6 0
2 years ago
Read 2 more answers
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