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3 years ago

What is the photoelectric effect?

Physics

2 answers:

Veronika [31]3 years ago

5 0

Answer:

emissions of electrons from a metal

Explanation:

I did grad point

n200080 [17]3 years ago

4 0

the emission of electrons or other free carriers when light shines on a material.

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A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

1 m=0.00062137 miles

1 hour=3600 s

7 0

3 years ago

An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

Monica [59]

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

6 0

2 years ago

I drove from Oklahoma to California (1452.9 miles) in 23 hours? What was my average speed?

storchak [24]

Answer:

63 miles per hour

Explanation:

you just divide miles and hours to get your average speed

5 0

2 years ago

A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of

Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0

3 years ago

In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kozerog [31]

Answer: vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s vf2 = 1.5 m/s

8 0

3 years ago