(a)
The formula is:
∑ F = Weight + T = mass * acceleration
as the elevator and lamp are moving downward, I choose downward forces to be
positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2
(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides
(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63
mass = 63 / 11.5
Mass = 5.48 kg
(b)
Since the elevator and lamp are going upward, I choose upward forces to be
positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7
-53.70 + T = 5.48 * 1.7
T = 53.70 + 9.316 = approx 63 N
The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated
upward at the same rate of 1.7 m/s^2
Via half-life equation we have:
Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:
There is 6.25 grams left of Ra-229 after 12 minutes.
Answer:
resolve shear stress = 22 MPa
Explanation:
Given data
slip plane α = 43.1°
slip directions β = 47.9°
shear stress = 20.7 MPa (3,000 psi)
applied stress =45 mPa (6,500 psi)
to find out
what stress will be necessary
solution
we know that
resolve shear stress = aplied stress × cosα × cosβ
resolve shear stress = 45 × cos(43.1) × cos(47.9)
resolve shear stress = 22 MPa
we can say that here single cristal will be yield
because resolve shear stress is bigger than critical shear stress
The amount of heat will be equal to Lm.
Where L is the latent heat of fusion and m is mass of the ice.
Latent heat of ice = 80cal/g.
So the amount of heat required here will be 35× 80cal
= 2,800 cal.
The basketballs and racquetball eventually stopped bouncing due to the first law. the first law states that an object at rest will stay at rest unless acted upon by an external force. this means that once the 2 balls lose friction, they will remain at rest until acted upon by an external force. once the 2 balls lose friction(energy), they come to a complete stop.
