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hjlf
3 years ago
8

The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl

ete the 1360 m run in 89.49 seconds. What is the acceleration of the sled?
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0
acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}
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If the frequency of q of a trait is 0.4, what is the frequency of p?
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A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.
natima [27]

Answer:

(a)  24.025 m/s. downward.

(b)  31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish  = 29.41+1.25

Distance between the pelican and fish  = 30.66

Distance between the pelican and fish ≈ 31 m

3 0
3 years ago
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