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givi [52]
2 years ago
9

When an automobile rounds a curve at high speed, the loading (weight distribution) on the wheels is markedly changed. For suffi-

ciently high speeds the loading on the inside wheels goes to zero, at which point the car starts to roll over. This tendency can be avoided by mounting a large spinning flywheel on the car. (a) In what direction should the flywheel be mounted, and what should be the sense of rotation, to help equalize the loading
Physics
1 answer:
NISA [10]2 years ago
4 0

Answer:

Explanation:

The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .

Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .

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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal
Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

6 0
2 years ago
A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?
lana [24]
KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps
7 0
2 years ago
Read 2 more answers
A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of
diamong [38]

Answer:

 2n t = m λ₀ ,    R = 0.240 mm

Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,

-This phase change is the introduction of a λ/2 change

-The ray passing through the film has a change in wavelength due to the refractive index of the medium

          λ₀ = λ / n

Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is

        2 t = (m’+ ½ + ½) λ₀ / n

        2t = (m’+1) λ₀ / n

         m = m’+ 1

        2n t = m λ₀

        With   m = 0, 1, 2, ...

Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength

The thickness of a hair is the thickness of the film t

           2R = t

             R = t / 2

             R = 0480/2

              R = 0.240 mm

3 0
1 year ago
The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.
melamori03 [73]
If you subscribe I’ll answer QF Aotrx
8 0
2 years ago
3. In 1989, Michel Menin of France walked on a tightrope suspended under a
Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0
2 years ago
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