Is the answer true or false?

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

Klio2033 [76]

Answer:

The radius of a chamber is 2.36 meters.

Explanation:

Given that,

The outer wall moves at a speed of 2.72 m/s.

Mass of the person, m = 75.1 kg

The person feels a force of 235 N force pressing against his back. The force acting on the person is centripetal force. It is given by the below formula :

$F=\dfrac{mv^2}{r}$

r is the radius of a chamber

$r=\dfrac{mv^2}{F}$

$r=\dfrac{75.1\times (2.72)^2}{235}$

r = 2.36 meters

So, the radius of a chamber is 2.36 meters. Hence, this is the required solution.

7 0

3 years ago

1 What conclusion can you draw about semicircular canals from their name ? A Their shape B Their size Their location in your bod

oee [108]

Answer:

A. Their shape

Explanation:

The name clearly shows that the shape of these canals are semi-circular. The semi-circular canals consist of three tubes filled with fluid and located in the inner ear. They help to maintain balance and transmit impulses through the movement of the fluids. The impulses sent through these fluids are sent to the brain for interpretation.

The function of the canals are not indicated by the name, rather the shape is hinted through the name.

7 0

3 years ago

A plastic rod is charged up by rubbing a wool cloth, and brought to an initially neutral metallic sphere that is insulated from

andrew-mc [135]

Answer:

Yes option A is right.

Explanation:

As we know that the "Opposite charges attract and like charges repel eachother". So based upon that fact we find out the sphere will be repelled or attract by the rod. As in this case metallic sphere was neutral initially but then we touched the rod with it. Although it was for few seconds but the charge is transferred to the sphere. Now both sphere and the rod have charge. After the seperation we look towards their respond If both have the opposite charge they will attract eachother. But here in this case they repel because they have the same charge, as we have charged the neutral sphere with the rod so we already know that they have the same charges that is why they are repelling eachother.

Insulation from the ground means that blocking the way of charges or free electrons from earth to metallic sphere and vice versa. As there exists free electrons and charges in earth they would flow into the metallic objects. So for more precise and accurate experiments we insulate the metals or prevent the metals from touching the earth surface to avoid the flow of charges through them. I hope it will help you.

5 0

3 years ago

A box with sides of length 40 cm is made up of six pieces of concrete of thickness 5 mm. The inside of the box is filled with 10

Damm [24]

Answer:

d)2.13 C s⁻¹

Explanation:

Rate of flow of heat through walls

= $\frac{KA(T_2-T_1)}{d}$

K = .33

A = 6 X .4 X .4 =0.96

T₂-T₁ = 30+40 = 70

d = 5 x 10⁻³

Put these data in the relation above

Rate of flow of heat

= $\frac{.33\times.96\times70}{5\times10^{-3}}$

= 4435.2 Js⁻¹

Specific heat of gas = 2.5 R = 20.785 J

Rise in temp = $\frac{4435.2}{100\times20.785}$

= 2.13 degree celsius.

4 0

3 years ago

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little

horrorfan [7]

Answer:

t_up / t_down = 6.83

Explanation:

Find:

Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.

Solution:

- Compute the velocity v_o at y_max:

v_i^2 = v_f^2 - 2*g*y_max

0 = v_o^2 - 2*g*y_max

v_o = sqrt (2*g*y_max)

- The total time spend by athlete above height y_max / 2 is:

y - y_o = v_o*t_up - 0.5*g*t^2_up

v_o = 0.5*g*t_up

- Equate two equations:

sqrt (2*g*y_max) = 0.5*g*t_up

t_up = 2*sqrt(2*g*y_max) / g

- The total time taken by athlete to reach height y_max / 2 from ground is:

y - y_o = v_o*t_down - 0.5*g*t^2_down

g*t_^2down - 2*v_o*t_down + y_max = 0

- Solve the quadratic and evaluate t_down:

t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g

Substitute for v_o = sqrt (2*g*y_max)

t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g

- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.

Hence,

t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g

t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)

- Compute the ratio t_up to t_down:

t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)

= 2*sqrt(2) / (sqrt(2) - 1)

= 6.83

8 0

3 years ago