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emmainna [20.7K]
4 years ago
4

Two identical metal bars are heated up until they are both glowing. One of them is "red hot" and the other is "blue hot." Which

one is hotter, the one that glows red or the one that glows blue?A) the red oneB) the blue oneC) We cannot tell without knowing more about the two bars.
Physics
1 answer:
lara [203]4 years ago
8 0

Answer:

B) the blue one

Explanation:

We can assimilate each metal bar to a black body. The peak wavelength of the radiation emitted by a blackbody is given by Wien's displacement law:

\lambda = \frac{b}{T} (1)

where

b is the Wien's displacement constant

T is the absolute temperature of the object

In this case, we have one object hotter and the other one colder. We see from (1) that the peak wavelength is inversely proportional to the temperature: therefore, the hotter object will have shorter peak wavelength, while the colder object will have longer peak wavelength.

Since red light has longer wavelength than blue light, we can conclude that the object that glows blue is hotter than the one that glows red.

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3 years ago
With what speed does the can move immediately after the collision? Answer in units of m/s.
Ratling [72]

Answer:

1.74 m/s

Explanation:

From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.

Therefore, using the formula below, we can calculate the speed of the can, V(can);

===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).

Since the question says the collision was elastic, we use the formula below

Slotting in the given values into the equation (1) above, we have;

1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).

Therefore, final velocity of the can= 2M1V1/M1+M2.

==> 2×2.7×1.1/ 2.7 + 0.72.

The velocity of the can after collision = 1.74 m/s

7 0
3 years ago
A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

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4 years ago
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Someone help me with letter d. and e. I’m supposed to solve for y.
Bumek [7]

Answer:

t=\frac{a}{\sqrt{1-\frac{y^2}{c^2}}}\\\frac{y^2}{c^2}=1-\frac{a^2}{t^2}\\y^2=\frac{c^2t^2-c^2a^2}{t^2}\\y=\frac{c}{t}\sqrt{t^2-a^2}

t=\frac{a}{\sqrt{1-\frac{v^2}{y^2}}}\\\frac{v^2}{y^2}=1-\frac{a^2}{t^2}\\\\\frac{v^2}{y^2}=\frac{t^2-a^2}{t^2}\\y^2=\frac{v^2t^2}{t^2-a^2}\\y=\frac{vt}{\sqrt{t^2-a^2}}

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