Answer:
Explanation:
net force on the skier = mg sin 39 - μ mg cos39
mg ( sin39 - μ cos39 )
= 73 x 9.8 ( .629 - .116)
= 367 N
impulse = net force x time = change in momentum .
= 367 x 5 = 1835 kg m /s
velocity of the skier after 5 s = 1835 / 73
= 25.13 m /s
b )
net force becomes zero
mg ( sin39 - μ cos39 ) = 0
μ = tan39
= .81
c )
net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s
so he will have speed of 25.13 m /s after 5 s .
Answer:
Explanation:
a ) work done by gravitational force
= mg sinθ ( d + .21)
Potential energy stored in compressed spring
= 1/2 k x²
= .5 x 431 x ( .21 )²
= 9.5
According to conservation of energy
mg sinθ ( d + .21) = 9.5
3.2 x 9.8 x sin 30( d + .21 ) = 9.5
d = 40 cm
b )
As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.
mg sin30 = kx
3.2 x 9.8 x .5 = 431 x
x = 3.63 cm
When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.
Although X-rays have many benefits, they can be dangerous to human health because they are short wavelength, high frequency, with photons sufficiently energetic to damage DNA molecules.
Okay so don't quote me on this but I believe the answer is A) I'm saying this because B and C make no sense. and you can't change the mass of something without changing it totally.
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = 
we substitute the values
v_f = 
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = 
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = 
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
