Answer:
100 degrees Celsius
Explanation:
Water starts to boil at 100 degrees celcius or 212 degrees fahrenheit.
Answer:
∑Fy = 0, because there is no movement, N = m*g*cos (omega)
Explanation:
We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.
If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
Stress = 4.67 * 10^-7 N/m²
Explanation:
Young's modulus of the material = Stress/Strain
Given
Young's modulus = 228 x 10^9 Pa
Stress = 106,483 Pa
Required
Strain
From the formula;
Strain = Stress/Young modulus
Strain = 106,483 /228 x 10^9
Stress = 4.67 * 10^-7 N/m²
Answer:
The net friction force is 8.01 N
Explanation:
Net friction force = mass of hockey puck × acceleration
From the equations of motion
v^2 = u^2 + 2as
v = 40 m/s
u = 0 m/s (puck was initially at rest)
s = 30 m
40^2 = 0^2 + 2×a×30
60a = 1600
a = 1600/60 = 26.7 m/s^2
The acceleration of the puck is 26.7 m/s^2
Net friction force = 0.3 × 26.7 = 8.01 N
