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malfutka [58]
3 years ago
15

What is the conversion factor between km/h and mi/h

Physics
1 answer:
Aloiza [94]3 years ago
5 0
<span> 1 mi = 1.609 km
so X mi/hr = 1.609 * X km/hr hope this helps!!</span>
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Which is an inertial reference frame (or at least a very good approximation of one)? Which is an inertial reference frame (or at
ioda

Answer:

A jet plane flying straight and at level at constant speed

Explanation:

     The<em> inertial frame </em>of reference is a frame of reference in which all <em>Newton law  is valid</em> ie Newton second law of motion and therefore newton first law of motion holds good. <em>The frame of reference does not accelerate.</em>

All the object that is in the frame of reference are at rest or moving with constant rectilinear motion with constant velocity unless acted upon by any force.

4 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
2 years ago
Cavendish used the attraction between large and small lead balls to measure an experimental value for the gravitational constant
irina1246 [14]

The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true

The correct answer is true

<h3>How do you define gravitational constant?</h3>

the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.

<h3> where is the strongest gravity is?</h3>

The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.

To know more about gravitational constant visit:

brainly.com/question/858421
#SPJ9

8 0
10 months ago
Where are overcurrent protective devices normally installed in a branch circuit?
love history [14]

Overcurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

V is the voltage,I is the current and R is the resistance

The vercurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

Thus, In a branch circuit, overcurrent safety devices are often located from where the conductors get their supply.

Learn more about resistance from here, refer to the link;

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5 0
1 year ago
A 6.00 g lead bullet traveling at420 m/s is stopped by a large
Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}

with m the mass and v the velocity.

K=529.2 J

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

Q=\frac{K}{2}= 264.6\,J

To know what the increase in temperature is, we should use specific heat of lead:

c=125.604 \frac{J}{kg\.K}

The equation that relates specific heat, change in temperature and mass is:

Q=cm\varDelta T

solving for \varDelta T:

\varDelta T=\frac{Q}{cm}=\frac{264.6}{(125.604)(6\times10^{-3})}

\varDelta T= 351.10\, K

4 0
2 years ago
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