A mass of 2000 kg. is raised 2.0 m in 10 seconds. What is the kinetic energy of the mass at this height?

Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

8 0

3 years ago

The whooping crane (Grus americana) is the tallest bird native to North America. It almost went extinct in the 1900s; in 1938, t

grin007 [14]

Answer:

The value is $a = 2.7183$

Explanation:

From the question we are told that

The relationship between the number of whooping cranes and the number of decades is $ln N = 2.85 + 0.039t$

The exponential relationship is $N = na^{kt}$

Now from the given equation we have that

$N = e^{2.85 + 0.039t}$

$N = 17.29 e^{0.039t}$

So comparing this equation obtained an the given exponential relationship we have that

$n = 17.29$

$k = 0.039$

$a = e = 2.7183$

$a = 2.7183$

7 0

2 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

3 years ago

What do you think that astronomers mean when they use the term observable universe?

iren2701 [21]

The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us.

Hope this helps :)

6 0

3 years ago

Which of the following occurs with both a cold front and a mountain breeze?

BigorU [14]

Answer:

b warm air rises

Explanation:

5 0

2 years ago