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lakkis [162]
3 years ago
7

Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob

er
Physics
1 answer:
Svetach [21]3 years ago
3 0
"Midnight" means looking away from the Sun. But in 6 months from April to October the earth goes halfway around the Sun. So midnight in April and midnight in October are exactly opposite directions.
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A 0.22 caliber handgun fires a 1.9g bullet at a velocity of 765m/s. Calculate the de Broglie wavelength of the bullet. Is the wa
shusha [124]

Answer:

  • de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ m
  • The value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.

Explanation:

Given;

mass of the bullet, m = 1.9 g = 0.0019 kg

velocity of the bullet, v = 765 m/s

de Broglie wavelength of the bullet is given by;

\lambda = \frac{h}{mv}

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

λ is de Broglie wavelength of the bullet

\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\  \lambda =4.56 *10^{-34} \ m

Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.

4 0
3 years ago
Question 21 of 25
sashaice [31]

Answer:

C. 1.3 x 10-11 F

I took the test and got this correct, hope this helps

Explanation:

6 0
3 years ago
Your body converts chemical energy to all of the following except
Musya8 [376]
NUCLEAR ENERGY BECAUSE THAT WOULD BE DANGEROUS. 

8 0
3 years ago
I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
Which statement is equivalent to newtons first law
larisa [96]

Newtons first law states "an object at rest will stay at rest while an object in motion will remain in motion" im confused as to what you are asking for but heres an example.

a car traveling at a speed of 35mph coming up to a stop sign has to slow down to a stop it cant stop instantly thats the same for it accelerating.

3 0
4 years ago
Read 2 more answers
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