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2 years ago

A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state.

1 answer:

8 0

The photon can be absorbed and the energy of the photon is exactly equal to the energy-level difference between the ground state and the level d.

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Determine the torque applied to the shaft of a car that transmits 225 hp

Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

$T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }$

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

$T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu$

8 0

2 years ago

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

choli [55]

Answer:

25 times the average speed

7 1

2 years ago

Read 2 more answers

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

2 years ago

If you wanted to measure the width of the gym, how would the accuracy of a meter stick compare with that of a 50m tape

Alisiya [41]

Meter stick would not be as accurate,
Every time you placed it down and picked it back up you run the chance of losing 2-4 cm each time.

7 0

2 years ago

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26× $10^{13}$

==> $v^{2}$ = 1.1 × $10^{19}$ ÷ 6.26× $10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5× $10^{8}$ ÷ 0.419188323 × $10^{3}$

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0

2 years ago

Read 2 more answers