9. A cube has a mass of 40g, a volume of 8cm' and a length of 2. What<br>is the density?

1) What do (x) and (y) symbolize?

MArishka [77]

Answer:

x is vertical and y is horizontal

Explanation:

7 0

3 years ago

I dropped an apple (mass 0.1kg) from the window because i'm weird. (15m above the ground). How fast was it going when it hit the

Olegator [25]

Answer:

I think the answer is 1 m per second.

5 0

3 years ago

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from the

Triss [41]

Answer:

Check the explanation

Explanation:

This is the step by step explanation to the above question:

$v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]$

= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]

v = 344 m/s

vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))

= 0.74 m/s

8 0

3 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

6 0

1 year 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:

<u>Knowns </u>

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)

<u>Given </u>

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

<u>Calculations</u>

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

Read 2 more answers

9. A cube has a mass of 40g, a volume of 8cm' and a length of 2. Whatis the density?​

9. A cube has a mass of 40g, a volume of 8cm' and a length of 2. Whatis the density?