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2 years ago

7

please help!!!!!what is the main function of a telescope? A.it brings scientists closer to distant obejects B. it magnifies ligh

t recieved from distant objects C. it measures the temperature of distant objects D.it measures the wavelengths of incoming light rays?

2 answers:

Illusion [34]2 years ago

8 0

The answer is B. Hope that helped.

Savatey [412]2 years ago

8 0

Answer is....

B. it magnifies light received from distant objects,

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A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magne

ser-zykov [4K]

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R = 16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

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3 years ago

The formula can be used to convert temperatures between degrees Fahrenheit () and degrees Celsius (). How many degrees are in th

Jlenok [28]

Answer:

-30°C

Explanation:

F-32/180 =C-0/100

or, -22-32/180=C/100

or, -54/180*100=C

or, -0.3*100=C

therefore, C= -30

-22°F = -30°C

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2 years ago

What's a sentence using work in a nonscientific sense

kvv77 [185]

My Physics teacher keeps giving us way too much work to do at home.

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3 years ago

Read 2 more answers

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago