All categories

3 years ago

It’s bungee jumping skydiving and hiking

Physics

1 answer:

butalik [34]3 years ago

4 0

That’s a great question

You might be interested in

A novice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north. if the wind blows at 75 km/

serg [7]

Velocity of plane is given as

$v_1 = 2.50 * 10^2 km/h$ towards north

$v_2 = 75 km/h$ towards south east

we can write the two velocity in vector form

$v_1 = 2.50 * 10^2 \hat j$

$v_2 = 75 cos45 \hat i - 75 sin45 \hat j$

now the net velocity will be given as

$v_{net} = v_1 + v_2$

$v_{net} = 250\hat j + 53.03 \hat i - 53.03 \hat j$

$v_{net} = 53.03\hat i + 196.97 \hat j$

so the resultant velocity will be given as

$v_{net} = \sqrt{53.03^2 + 196.97^2}$

$v_{net} = 203.98 m/s$

and the angle of velocity will be

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{196.97}{53.03} = 74.9 degree$

3 0

4 years ago

Which of the following objects conducts heat the best?

JulijaS [17]

I think its C. because metal is a good conductor of heat

I hope this helps ^-^

let me know if im wrong

4 0

4 years ago

Read 2 more answers

Cars with hybrid and electric motors what?

kogti [31]

Are silent run on power and have to be electricly charged

5 0

3 years ago

Need help with this really lost<br> A<br> B<br> C<br> D

djverab [1.8K]

The answer is "C" because the equator is closest to the sun, because the sun's rays hits the earth surface at the higher angel at the equator and also the sun is directly overhead the equator.

7 0

4 years ago

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

madreJ [45]

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is $E^o_{cell} = 2.35 V$

Explanation:

From the question we are told that

The ionic equation is

$3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}$

Now under standard conditions the reduction half reaction is

$Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V$

And the oxidation half reaction is

$Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V$

The emf of this cell under standard conditions is mathematically represented as

$E^o_{cell} = E^o _r - E^o _o$

substituting values

$E^o_{cell} = 1.61 - (- 0.74)$

$E^o_{cell} = 2.35 V$

3 0

3 years ago