Ask question

Ask question

All categories

3 years ago

9

Which statement about a Ferris wheel is true?

1 answer:

nikitadnepr [17]3 years ago

6 0

Answer:the imput force is applied to the axle

Explanation:

You might be interested in

PLEASE HELP!!!!!

never [62]

Answer:

The answer is A

Explanation:

When a rockets thrusters push on the ground the ground pushes back on the rocket with equal force in the opposite direction. Hence the rocket takes off.

Newtons third law of motion states, for every action there is an equal and opposite reaction.

7 0

2 years ago

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

5 0

3 years ago

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to ro

Dafna11 [192]

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

$T_x = nsinA = \frac{mv^2}{r}$

$T_y = ncosA = mg$

Dividing both.

$tan A = \frac{v^2}{rg}$

$tan A = \frac{11.7^2}{50*9.8}$

$A = tan^{-1} (0.279367)$

$A = 15.608\°$

Therefore the angle that should the curve be banked is 15.608°

7 0

4 years ago

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

sesenic [268]

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

$W_1 = 12*5cos15$

$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

$W_2= -20 J$

Now total work done on sled

$W_{net}= W_1 + W2$

$W_{net} = 57.96 - 20 = 37.96 J$

7 0

3 years ago

If you move 50 meters in 10 seconds, what is your speed

iogann1982 [59]

Explanation:

My answer didn't save :(

8 0

3 years ago