The equation below is used to calculate the mechanical advantage of an ideal wheel and axle.

What is the mass of an object travelling at 25 m/s with a kinetic energy of 3775 J?

Arte-miy333 [17]

Answer:

<h2>You can do 6516-3775=2741</h2><h2 /><h2>The difference is 27141 kilometers</h2><h2 /><h2>Did this help?</h2><h2>⇒ Yes or No?</h2>

4 0

2 years ago

What causes a solar eclipse? Select all that apply.

brilliants [131]

I think it is either D) or E)

But i am going to go with E)

4 0

2 years ago

Read 2 more answers

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

elena55 [62]

Answer:

$3.2\times 10^{-7}\ m$ or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = $9.38\times 10^{14}\ Hz$

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = $3\times 10^8\ m/s$

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

$v=f\lambda$

Now, expressing the above equation in terms of wavelength 'λ', we have:

$\lambda=\frac{v}{f}$

Now, plug in the given values and solve for 'λ'. This gives,

$\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m$

Therefore, the wavelength of the radiations absorbed by the ozone is nearly $3.2\times 10^{-7}\ m$ or 0.32 μm.

7 0

2 years ago

An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

Alekssandra [29.7K]

Answer:

$0.53m/s^2$

Explanation:

We are given that

Mass of wire=m=3.6 g= $3.6\times 10^{-3} kg$

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms= $60.1\times 10^{-3} s$

$1 ms=10^{-3} s$

Speed,v= $\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s$

$g=\frac{v^2m}{m'l}$

Using the formula

$g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2$

8 0

3 years ago

6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if it

Salsk061 [2.6K]

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

$\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\$

The flux is given by

$\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C$

3 0

2 years ago