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jekas [21]
3 years ago
12

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

a 1900n force on the road and produces an acceleration of 0.550 m/s2? the mass of the boat plus trailer trailer is 700kg. (b) what is the force in the hitch betweenthe car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

Physics
1 answer:
likoan [24]3 years ago
3 0
Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g,  the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where  =  kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

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y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s

Part (a)

To find the final velocity we use the kinematic equation

So

(v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s

Part (b)

To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation

v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds

Notice that we substituted vf with negative sign because its direction is downwards

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22. What is the length of a pendulum that has a period of 0.500 s?
sergeinik [125]

6.21 cm

  • use the pendulum formula : \sf \bold{\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}}

where

  • T is time or period
  • π is pie = 22/7
  • L is pendulum length
  • g is acceleration due to gravity

Given:

  • T = 0.500 s
  • g = 9.8 m/s²

solving step-wise:

\dashrightarrow \mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{g}}}

\sf \dashrightarrow \mathrm{0.5}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}

\dashrightarrow \mathrm{\dfrac{0.5}{2 \pi } }=\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}

\dashrightarrow\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= \mathrm{\dfrac{0.5}{2 \pi } }

\sf \dashrightarrow{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2

\sf \dashrightarrow{{\mathrm{L}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2*9.8

\sf \dashrightarrow{{\mathrm{L}}=0.06205922 \ m

  • 1 m → 100 cm

\sf \dashrightarrow{{\mathrm{L}}=6.2059\ cm

\sf \dashrightarrow{{\mathrm{L}}=6.21\ cm             { rounded to nearest hundredth }

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2 years ago
Read 2 more answers
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