In the picture above, what is the period of the pendulum?<br> (Remember to include a unit)

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi

miv72 [106K]

Answer:

The time taken is $\Delta t = 1.5 \ s$

Explanation:

From the question we are told that

The mass of the ball is $m = 1.25 \ kg$

The time taken to make the first complete revolution is t= 3.60 s

The displacement of the first complete revolution is $\theta = 1 rev = 2 \pi \ radian$

Generally the displacement for one complete revolution is mathematically represented as

$\theta = w_i t + \frac{1}{2} * \alpha * t^2$

Now given that the stone started from rest $w_i = 0 \ rad / s$

$2 \pi =0 + 0.5* \alpha *(3.60)^2$

$\alpha = 0.9698 \ s$

Now the displacement for two complete revolution is

$\theta_2 = 2 * 2\pi$

$\theta_2 = 4\pi$

Generally the displacement for two complete revolution is mathematically represented as

$4 \pi = 0 + 0.5 * 0.9698 * t^2$

=> $t^2 = 25.9187$

=> $t= 5.1 \ s$

So

The time taken to complete the next oscillation is mathematically evaluated as

$\Delta t = t_2 - t$

substituting values

$\Delta t = 5.1 - 3.60$

$\Delta t = 1.5 \ s$

7 0

3 years ago

Determine the mass the mass in Kilograms of a Object that has a weight of

34kurt

1. 20mn that’s the answer

6 0

2 years ago

Read 2 more answers

What is produced as the result of unequal warming of the earth's surface?

Soloha48 [4]

Unusual precipitation patterns

7 0

3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

4 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$