Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
Answer:
change in y = -7
change in x = -17
magnitude of displacement = sqrt(7^2+17^2)
tan of angle below -x axis = 7/17
because in third quadrant where x and y are negative
Answer:
See the answers below.
Explanation:
We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F =m*a
where:
F = force [N] (units of newtons)
m = mass = 1000 [kg]
a = acceleration = 3 [m/s²]
![F = 1000*3\\F=3000[N]](https://tex.z-dn.net/?f=F%20%3D%201000%2A3%5C%5CF%3D3000%5BN%5D)
And the weight of any body can be calculated by means of the mass product by gravitational acceleration.
![W=m*g\\W=1000*9.81\\W=9810 [N]](https://tex.z-dn.net/?f=W%3Dm%2Ag%5C%5CW%3D1000%2A9.81%5C%5CW%3D9810%20%5BN%5D)
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
The distance is 30 km and the displacement is 22.4 km North East
