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3 years ago

Which statement about protons is true?

2 answers:

8 0

Answer to your question is A. Because you cannot have HALF of a Proton...
So of course you will always have a whole number...

White raven [17]3 years ago

4 0

GIVE vo0Mac0ov Brainliest ... He's smart

You might be interested in

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

3 years ago

A car travels at a constant velocity of 20.0 meters/second for 15 seconds. What is the power of the car if the initial force app

Zielflug [23.3K]

Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J. {S=ut because acceleration is zero since car is moving at constant velocity}.

3 0

3 years ago

Read 2 more answers

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

4 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination

kondor19780726 [428]

The frictional force is directly proportional to the force that is perpendicular on the surface.

When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction

Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

7 0

3 years ago