The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
Answer:

Explanation:
From the question we are told that:
Height 
Time 
Generally the Newton's equation for Initial velocity upward is mathematically given by



Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28


Generally the Newton's equation for time to reach initial velocity is mathematically given by




Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Answer:
m=146.277kg which is rounded to 146kg
Explanation:
Remember that F=ma
But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.
So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg
Mass is always in kg, unless stated otherwise.