The quantity represented by vi is a function of time (i.e., is not constant).A. TrueB. False

A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules

kobusy [5.1K]

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

3 0

3 years ago

At what water temperature will additional heat energy need to be added before the temperature will change again?

Lostsunrise [7]

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)<span />

6 0

2 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

2 years ago

Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a

timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

$a_2 = 2.12 m/s^2$

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

$F = 0 = m_1a_1 + m_2a_2$

now we have

$m_1 = 0.200 kg$

$m_2 = 0.250 kg$

$a_1 = 2.65 m/s^2$

now we have

$0.200(2.65) + 0.250a_2 = 0$

so we have

$a_2 = 2.12 m/s^2$

8 0

3 years ago

A transverse wave on a long horizontal rope with a

frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and speed by the equation

$f=\frac{v}{\lambda}$

where

f is the frequency

v is the speed of the wave

$\lambda$ is the wavelength

For the wave in this problem,

v = 2 m/s

$\lambda=8 m$

So the frequency is

$f=\frac{2}{8}=0.25 Hz$

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

$T=\frac{1}{f}=\frac{1}{0.25}=4 s$

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

$t=\frac{T}{2}=\frac{4}{2}=2 s$

4 0

2 years ago