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andreev551 [17]
3 years ago
13

G davisson and germer performed their experiment with a nickel target for several electron bombarding energies. at what angles w

ould they find diffraction maxima for 48 ev and 60 ev electrons?
Physics
1 answer:
Anton [14]3 years ago
5 0
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
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I MARK BRAINLIEST, PLEASE ANSWER ASAP!!! PHYSICS 20 POINTS!!
irina [24]
The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!
5 0
3 years ago
Read 2 more answers
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
A person on a merry go round is constantly changing direction
Salsk061 [2.6K]

Answer: yurp

Explanation:

because its spinning

7 0
3 years ago
Read 2 more answers
a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her
Rama09 [41]
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
5 0
3 years ago
Read 2 more answers
Use the following half-life graph to answer the following question:
Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

m_0 = 50.0 g

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

m'=\frac{50.0 g}{2}=25.0 g

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0
3 years ago
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