Ask question

Ask question

All categories

3 years ago

5

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the

diver, in rad/s, during this time interval.

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

The angular speed is $w = 5.89 \ rad/s$

Explanation:

From the question we are told that

The time taken is $t = 1.6 s$

The number of somersaults is n = 1.5

The total angular displacement during the somersault is mathematically represented as

$\theta = n * 2 * \pi$

substituting values

$\theta = 1.5 * 2 * 3.142$

$\theta = 9.426 \ rad$

The angular speed is mathematically represented as

$w = \frac{\theta }{t}$

substituting values

$w = \frac{9.426}{1.6}$

$w = 5.89 \ rad/s$

You might be interested in

What is the SI unit for moment

natali 33 [55]

Answer:

Newton Metre

Explanation:

8 0

3 years ago

You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that

kogti [31]

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

6 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

Heres some earned points

egoroff_w [7]

Answer:

what do you mean by that.

5 0

2 years ago

Read 2 more answers

The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount

igor_vitrenko [27]

In order to decrease the friction on the slide,
we could try some of these:

-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide. The oil oozes
down the slide and keeps the whole slide greased.

-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.

-- Spray the whole slide with soapy sudsy water, every 30 minutes.

-- Drill a million holes in the slide,and pump high-pressure air
through the holes. Make the slide like an air hockey table.

-- Keep the slide very cold, and keep spraying it with a fine mist
of water. The water freezes, and a thin coating of ice stays on
the slide.

-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide. If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.

-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide. Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.

7 0

3 years ago