Answer:
Acceleration will be change at every instant of time
Explanation:
We have given that speed of the car at curve round is constant
We know that speed is scalier quantity which does nit depend on the direction of motion.
Velocity is vector quantity which depends on the direction of motion
As on curve round direction of motion is always changes so velocity is also changes and acceleration is rate of change of velocity
So acceleration of the car will also change.
Answer:
Explanation:
While each of these can cause erosion and weathering, lightning is probably the least important as it occurs less frequently and affects a much smaller surface area when it strikes.
Wind is not very effective by itself, but it can carry abrasives which work to degrade rock surfaces. It covers a very large area at once so the net effect can be moderate to large especially desert areas where plants are not readily available to disrupt the flow.
Rain covers huge areas and is quite common.
Freezing/Thawing cycles cover large areas and are quite common in the temperate and arctic latitudes and even in tropical altitudes.
Attached is a photo taken atop Half Dome in Yosemite National Park showing two of thousands of divots in the rock there caused by lightning strikes. The current in the lightning heats the stone causing water trapped in it to flash to steam. The increased pressure inside the stone can overwhelm the material strength and blow rock chunks over a fairly good sized area. This is a fairly rapid weathering and erosion when it occurs, but that is typically limited to a few dozen days per year and occurs mostly on high ground where lightning is more likely to strike earth.
The value of normal force as the slider passes point B is
The value of h when the normal force is zero
<h3>How to solve for the normal force</h3>
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
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The answer would be D, negative acceleration
Answer:
d = 146.2 m
increase
Explanation:
Using Energy Balance:
U_i + W = U_f
K_i + U_gi + W = K_f + U_fg
0 + m*g*h + F*d = 0.5*m*v_f^2
(900)*(9.81)*(50 sin (5)) + 400*50 = 0.5*900*v_f^2
v_f^2 = 129.944
v_f = 11.4 m/s at the bottom of the hill
U_i + W = U_f
K_i + U_gi + W = K_f + U_fg
0.5*m*v_i^2 - F*d = 0
0.5*900*129.944 = 400*d
d = 146.19 m
part b
Heavier car will gain greater momentum at the bottom of the hill and attain a higher velocity at the bottom; hence, the distance will increase.
Answer: increase