45✖️77 and then you will get your answer
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
Answer:
a) Andrea's initial momentum, 200 kg m/s
b) Andrea's final momentum, 0
c) Impulse, = - 200 Ns
d) The force that the seat belt exerts on Andrea, - 400 N
Explanation:
Given data,
The initial velocity of the car is, u = 40 m/s
The mass of Andrea, m = 50 kg
The time period of deceleration, a = 0.5 s
The final velocity of the car, v = 0
a) Andrea's initial momentum,
p = mu
= 50 x 40
= 200 kg m/s
b) Andrea's final momentum
P = mv
= 50 x 0
= 0 kg m/s
c) Impulse
I = mv - mu
= 0 - 200
= - 200 Ns
The negative sign indicated that the momentum is decreased.
d) The force that the seat belt exerts on Andrea
F = (mv - mu)t
= (0 - 200) / 0.5
= - 400 Ns
Hence,the force that the seat belt exerts on Andrea is, - 400 N
Answer:
Explanation:
Given
mass of cannon and cart is 
Spring constant 
mass of Projectile 
Launch Velocity of cannon 
Launch angle 
As the External Force is zero therefore we can conserve momentum
Initially both Projectile and cannon is at rest
Conserving momentum in horizontal direction
i.e. cannon is moving in opposite direction of Projectile
In my opinion,I think the answer is b.You can control variables more easily by doing different things for different purposes.
