Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
the outermost layer of Earth’s lithosphere that
is found under the oceans and
molded at scattering
centres ono
ceanic ridges, which occur at deviating plate boundaries
Oceanic crust is about 6 km (4 miles) thick.
hope it helps
Answer:
Answer:
New speed of the 22-kg block is 1.57 m/s
Explanation:
Mass of block
Mass of another block
Initial speed of the block
Initial speed of the another block
Initial speed of the another block
For conservation of momentum, we have
Substitute all the values and solving for final speed of the 22kg block is
new speed of the 22-kg block is 1.57 m/s
Couldnt write the answer so check picture
Answer:
Explanation:
Givens
Heat of Fusion = 2.05 * 10^5 J / kg watch the units.
Heat to actually melt the copper = 82 10^5 J
Formula
Mass of copper = Heat / Heat of Fusion
Solution
Mass of copper = 82*10^5 J / (2.05 * 10^5 J / kg)
Mass of copper = 40 kg
Notice that the kg is in the denominator of the second fraction. The rules of fractions would tell you the 1/1 / / 1 /kg . You take the right fraction and turn it upside down and multiply. 1 / 1 * kg/1 = 1* kg / 1*1 which is just kg.
Answer 40 kg of copper
Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
