Where do we use friction in day to day life

Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act

adoni [48]

Answer:

Juno scientific payload includes:

A gravity/radio science system (Gravity Science)
A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
A vector magnetometer (MAG)
Plasma and energetic particle detectors (JADE and JEDI)
A radio/plasma wave experiment (Waves)
An ultraviolet imager/spectrometer (UVS)
An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

3 0

2 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration

kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v= $\sqrt{2gh}$