Two difference between hydraulic brake and hydraulic lift

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. the maximum speed of the o

Andreas93 [3]

First of all, the maximum speed occurs when the object passes through the
equilibrium position

The kinetic energy when the object has this max speed is

K= 1/2 * mass * (1.25 m/s)^2

The potential energy in the spring when the speed is equal to zero

U= 1/2 * k * xmax^2

The maximun force of the spring is

mass*acceleration = k*xmax

m * 6.89 m/s2 = k * xmax
xmax = 6.89* m / k

0.5 * m * 1.56 = 0.5 * k * xmax^2

m * 1.56 = k * (6.89* m / k )^2 

1.56 m = 47.47 m^2 / k
m/k = 0.032862

period = 2 *pi*sqrt[m/k]
= 2 pi sqrt [ 0.032862]
= 1.139 s

A fourth of the period elapses between the instants of max acceleration and maximum speed

= 1/4* period
= 1/4 * 1.139 s = 0.284s

7 0

4 years ago

A body builder exerts 15 N of force over 3 m. How much work did she do?

tangare [24]

Work done = force x distance

work done = 15 x 3

work done = 45J

4 0

3 years ago

Read 2 more answers

Does the shape of a skateboard affect how will travel?

nika2105 [10]

Answer:

Yes

Explanation:

The shape will affect the friction and the way the skateboard moves.

6 0

3 years ago

Two small nonconducting spheres have a total charge of 90.0 C.

valentina_108 [34]

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

8 0

3 years ago

Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples.

Angelina_Jolie [31]

In scientific terms, ultrasound is a sound pressure, cyclic in nature, that has a greater frequency than the limit at the top of human hearing capabilities. What this means is that an ultrasonic sound can’t be heard by the human ear because their frequency is too high for our ears to pick up. In healthy young adults, this upper hearing capability is an average of 20 kilohertz. Ultrasound has many applications in several fields. Perhaps the best known application for ultrasound is sonography. This is where medical staff use the high pitched noise to produce a picture of a fetus while in the mother’s womb. Another use however, doesn’t directly concern humans at all. Bats use the high pitched noises to see in the dark and get an accurate reading on their preys internal structure. A popular belief is that an ultrasonic sound has the ability to turn the locking mechanism in a door lock, as demonstrated on some spy movies. On the opposite side of this are infrasonic sounds. These are noises with a frequency less than the lowest level of human hearing capabilities is 20 hertz. It is possible for humans to perceive infrasonic sounds, but only if the air pressure is sufficient. Although the war is the main tool for hearing these low sounds, it is possible for other parts of the body to “feel them”. Infrasound can be used to send signals in the army to special machines that can pick them up. These can be used to transmit vital data. Animals are able to pick up some low infrasonic noises which warn them of natural disasters before they happen, generally earthquakes and tsunamis.

I hope some of this information I gave you can help you. I came up with everything myself to help you.

4 0

4 years ago