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Katyanochek1 [597]

4 years ago

6

A ball is tossed vertically upwards with a speed of 5.0 m s–1. After how many seconds will the ball return to its initial positi

on?

1 answer:

mezya [45]4 years ago

7 0

Answer:

1.02 seconds

Explanation:

Initial velocity = u = 5 m/s

t = Time taken

s = Displacement travelled = 0

a = Acceleration due to gravity = -9.81 m/s² (negative sign due to direction)

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow 0=5t+\frac{1}{2}-9.81\times t^2\\\Rightarrow 0=5t-4.905t^2\\\Rightarrow 0=t(5-4.905t)\\\Rightarro 0=5-4.905t\\\Rightarrow t=\frac{-5}{-4.905}=1.02\ seconds$

So, time taken to return to its initial position is 1.02 seconds

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A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
$F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

For more information on this visit

brainly.com/question/21811998

5 0

3 years ago

What is a good hypothesis when putting pop rocks into soda?

8_murik_8 [283]

That it will erupt upon contact. Hope it helps!

6 0

3 years ago

Today chemistry is based on the law ofwhat is the matter

melisa1 [442]

Answer:

One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant.

Explanation:

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5 0

3 years ago

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

uranmaximum [27]

Answer:

The magnitude of force exerted on the handle is 108.73 N

Explanation:

To determine the magnitude of force exerted, we will use the formula relating Power and Force.

Power is the rate at which work is done. Power can be calculated from the formula

Power = Work / Time

But, Work = Force × Distance

Hence,

Power is given by the formula

$P = \frac{F \times s}{t}$

Where P is the Power

F is the force

s is the distance

and t is the time

From $P = \frac{F \times s}{t}$ ,

Then we can write that

$F = \frac{P \times t}{s}$

From the question,

Distance, s = 1.1 m

Time, t = 1.3 s

Power, P = 92 W

Putting these values into the formula, we get

$F = \frac{92 \times 1.3}{1.1}$

$F = \frac{119.6}{1.1}$

$F = 108.73N$

Hence, the magnitude of force exerted on the handle is 108.73 N.

7 0

3 years ago

At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug

Rom4ik [11]

- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

$\tan \theta=\frac{6}{x}$

Now, differentiate with respect to t, we get

$\sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}$

Now, calculate the travel distance from radar station to plane after 12min

Distance, $x=800 \times \frac{12}{60}=160$

Substituting ‘x’ value, we get

$\tan \theta=\frac{6}{160}=\frac{3}{80}$

Find the rate of change of theta after 12 min,

$\frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}$

We know, the formula for,

$\sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}$

So, then, $\frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })$

$\frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800$

$=-\frac{1}{1.0014} \times \frac{6}{25600} \times 800$

$=-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}$

When express the value in km/he, we get, the change in theta as

$=-187.237 \mathrm{km} / \mathrm{hr}$

5 0

4 years ago