The component of weight(
mg) that is responsible for the motion of boxes on ramp is:
Where m = mass of the boxes.
g = Acceleration due to gravity = 9.8
= The angle the ramp makes with the ground. In this case it is 30°.
Since the frictional force is:
μ*N.
Where,
μ = Frictional Coefficient
N = Normal to the ramp =
Therefore, the frictional force becomes =
= μ*
Apply Newton's second law we would get:
- μ*
=
=>
=
μ
-- (A)
Now according to equation of motion:
Where x = 5.8m
= 0
= 0
= 10.24
Plug in the value in the above equation you would get:
= 1.1328
Plug in
in equation (A) and solve for
μ, you would get,
μ = 0.4439
Answer:
1.86 s
Explanation:
Given the expression
h(t) = -16t²+ 64...................... Equation 1
Where h = height of the object, t = time it will take the object to hit the ground.
Given: h = 64 foot.
We have to concert from foot to meters
If 1 foot = 0.3048 meters
Then, 64 foot = 0.3048×64 = 19.51 meters.
We substitute the value of h into equation
119.51 = -16t²+64
-16t² = 199.51-64
-16t² = 55.51
t² = 55.51/-16
t² = 3.469
t = √3.469
t = 1.86 s.
Hence it will take the object 1.86 s to hit the ground.
First there are many ways to approach this problem. The simplest and most direct in my opinion is this approach:
vf^2=v1^2+2ad
so when the height reaches a maximum of 35m the velocity is zero for a split second before starting to fall. So we set vf=0 and solve for v1.
v1=sqrt(2gd)=sqrt(2*9.8*35)= 26.2m/s
for initial velocity needed to reach 35m height.
Any questions please ask!
Answer:
a) 378Ns
b) 477.27N
Explanation:
Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula
I = Ft = m(v-u)
m is the mass = 70kg
v is the final velocity = 5.4m/s
u is the initial velocity = 0m/s
Get the impulse
I = m(v-u)
I = 70(5.4-0)
I = 70(5.4)
I = 378Ns
b) Average total force is expressed as
F = ma (Newton's second law)
F = m(v-u)/t
F = 378/0.792
F = 477.27N
Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N
Answer:
A. Endothermic reaction.
B. +150KJ.
C. 250KJ.
Explanation:
A. The graph represents endothermic reaction because the heat of the product is higher than the heat of the reactant.
B. Determination of the enthalpy change, ΔH for the reaction. This can be obtained as follow:
Heat of reactant (Hr) = 50KJ
Heat of product (Hp) = 200KJ
Enthalphy change (ΔH) =..?
Enthalphy change = Heat of product – Heat of reactant.
ΔH = Hp – Hr
ΔH = 200 – 50
ΔH = +150KJ
Therefore, the enthalphy change for the reaction is +150KJ
C. The activation energy for the reaction is the energy at the peak of the diagram.
From the diagram, the activation energy is 250KJ.