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givi [52]
4 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]4 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 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
4 years ago
What mass, in grams, of aluminum fins could 2571 J of energy heat from 15.73 ∘C to 26.50 ∘C ? Aluminum has a specific heat of 0.
nadya68 [22]

Answer:

266 g or 0.266 kg

Explanation:

The formula for specific heat capacity is given as,

Q = cm(t₂-t₁) ..................... Equation 1

Where Q = Heat Energy, c = specific heat capacity of Aluminum, m = mass of the aluminum fins, t₁ = initial temperature, t₂ = final temperature.

make m the subject of the equation,

m = Q/c(t₂-t₁)................... Equation 2

Given : Q = 2571 J, c = 0.897 J/g.°C, t₁ = 15.73 °C, t₂ = 26.50 °C.

Substitute into equation 2

m = 2571/[0.897×(26.5-15.73)]

m = 2571/9.661

m = 266 g or 0.266 kg

Hence the mass of the Aluminum fins = 266 g or 0.266 kg

5 0
4 years ago
A pulley system has an efficiency of 87.5 percent.
Gelneren [198K]

Answer:

work done on desk = m g h = 105 * 9.81 * 2.46 = 2534 Joules

Explanation:

so work in = 2534 / 0.875 = 2896 Joules

that is your 648 times distance your hand moved holding the rope

2896 = 648 * x

4.47 meters

I think maybe a typo, 648 N is too big, maybe 64.8 ? Any block and tackle system does better than that.

7 0
3 years ago
Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e
anyanavicka [17]

Answer:

 Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

                F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

          F + F = m a

         k₃ x + k₃ x = m a

         a = 2k₃  x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

       Keq = 2k₃

6 0
3 years ago
Two coils, held in fixed positions, have a mutual inductance of M = 0.0034 H. The current in the first coil is I(t) = I0sin(ωt),
tiny-mole [99]

Answer:

ε₂ =2.63 V

Explanation:

given,

M = 0.0034 H

I (t) = I₀ sin (ωt)

I (t) = 5.4 sin (143 t)

\dfrac{d i(t)}{dt} = \dfrac{d}{dt}(5.4 sin (143 t))

\dfrac{d i(t)}{dt} =143 \times 5.4 cos (143 t)

magnitude of the induced emf in the second coil

ε₂ =M\dfrac{di}{dt}

ε₂ =0.0034\times 143 \times 5.4 cos (143 t)

for maximum emf

cos (143 t) = 1

ε₂ =0.0034\times 143 \times 5.4

ε₂ =2.63 V

3 0
3 years ago
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