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2 years ago

cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?

Physics

1 answer:

ikadub [295]2 years ago

5 0

Answer:

Displacement is 3 km North

Explanation:

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2 years ago

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As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.

scZoUnD [109]

Answer:22.36 m/s

Explanation:

Given

the diameter of loop d=50 m

the radius of loop r=25 m

At the top position, we can write,

weight and Normal reaction combination will provide the centripetal force i.e.

$R+W=\frac{mv^2}{r}$

$R=W\quad \quad [\text{apparent weight =Actual weight}]$

$2W=2mg=\frac{mv^2}{r}$

$v=\sqrt{2gr}$

$v=\sqrt{2\times 10\times 25}$

$v=22.36\ m/s$

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2 years ago

A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?

Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

$T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g} \ be \ constant \\\\\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\$

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

$\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\ T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\ T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s$

Therefore, its period if its length is increased by a factor of four is 5 s.

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2 years ago