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4 years ago

A 77.0 kg firefighter climbs a flight of stairs 8.0 m high. how much work is required? 0 incorrect: your answer is incorrect. j

1 answer:

5 0

<span>6160 joules

to lift 1 newton 1 metre requires 1 joule
there are 10 newtons in one kilo

so 77(kg) x 8 (metres) x 10 (newtons/kilo) = 6160 joules</span>

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In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

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3 years ago

What is the process called when earth's surface is broken down into smaller pieces

fiasKO [112]

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6 0

3 years ago

When in orbit, astronauts experience weightlessness what is this caused by?

erica [24]

no gravity or dark matter

4 0

3 years ago

Read 2 more answers

If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is

Law Incorporation [45]

Answer:

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

3 0

2 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago