As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.

8 0

3 years ago

If we increase the speed of rotation of wheel then the sound produced by cardboard touching the spokes of the wheel will become

Anarel [89]

it will become shrill

5 0

3 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

Read 2 more answers

The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s

Bumek [7]

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: $m_{(t)} =m_{0} e^{-kt}$ , where $m_{0}$ is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

$m_{(1)} =m_{0} e^{-k(1)}$ , but $m_{(1)}=\frac{99m_{0} }{100}$ , so we have $\frac{99m_{0} }{100}=m_{0}e^{-k}$ , then $k=-ln(\frac{99}{100})=0.01$

Now using k found we must to find $m_{(5)}$ .

$m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg$

6 0

3 years ago