state the type of force which provide the needed centripetal force in the following instance :an electron orbiting the nucleus

harkovskaia [24]

The answer is D
Neon has 20.0026 element more then the other

8 0

2 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

3 years ago

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

$Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz$

8 0

3 years ago

Which is the smallest unit of life?

emmasim [6.3K]

Answer:

the cell is the smallest unit

7 0

2 years ago

A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient

Basile [38]

Answer:

v_o = 4.54 m/s

Explanation:

Knowns

From equation, the work done on an object by a constant force F is given by:

W = (F cos Ф)S (1)

Where S is the displacement and Ф is the angle between the force and the displacement.

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:

K.E=1/2m*v^2 (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:

W = K.E_f-K.E_o (3)

Given

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020

Calculations

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:

∑F_y=N-mg

N=mg

Thus, the kinetic friction force is:

f_k = μ_k*N

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:

W_f=(f_k*cos(180) s

=-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:

W= -K.E_o

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force

W_f= -K.E_o

From equation (2), the work done by the friction force in terms of the initial speed is:

W_f=-1/2m*v^2

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:

-μ_k*mg*s = -1/2m*v^2

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s

v_o = 4.54 m/s

6 0

3 years ago

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