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klasskru [66]
4 years ago
13

What was the experiment carried out by Hershel & Ritter.

Physics
1 answer:
just olya [345]4 years ago
3 0
Hershel discovered something called Infrared light. It was basically an experiment outdoors and you have a light and when the light hits another surface (like a deck) then that surface heats up 
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At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s pe
KonstantinChe [14]

Answer:

4347.8 m/s  

Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB\\

Take the ratio:

m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s

3 0
3 years ago
URGENT HELP! <br> How is it that two moving objects can have a net momentum of zero?
Zielflug [23.3K]

Answer:

because they may not actually be moving be just percieved as so

Explanation:

8 0
3 years ago
Read 2 more answers
Which one of the following is correct? *
slamgirl [31]

Answer:

uniform acceleration motion is a motion with constant acceleration.

6 0
3 years ago
4. Which of the following would be a good reference point to describe the motion of a dog?
saul85 [17]

ANOTHER RUNNING DOG

Explanation:

In the given question it is to find a suitable reference point to describe  the motion of dog. Here I could suggest that it is better to compare the dog with  another running dog to create the relative speed difference to get a reliable motion variation.

Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to  another dog which is already in motion.

Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with  another dog running.

4 0
3 years ago
A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h
Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is m_1=1.6\times 10^4 kg

speed of bus towards North v_1=15.2 km/h\approx 4.22\ m/s

mass of bus travelling in South direction is m_2=1.578\times 10^4 kg

speed of bus v_2=12.2 km/h\approx 3.38\ m/s

mass of each Passenger in south moving bus m_0=64.8 kg

Momentum of North moving bus

P_1=m_1\times v_1

P_1=1.6\times 10^4\times 4.22

P_1=6.768\times 10^4 kg-m/s

Momentum with south moving bus

P_2=m_2\times v_2+n\cdot m_0\times v_2

P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38

For total momentum to be towards south

P_2-P_1 should be greater than 0

thus for least value of n

P_2=P_1

(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4

1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4

n=\frac{4243.6686}{64.8}=65.48\approx 66    

5 0
3 years ago
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