A basketball with a mass of 0.5 kilograms is accelerated at 2

PLEASE ASAP ILL GIVE BRAINLIEST.

Fiesta28 [93]

Acceleration no longer exist as the car stops.

5 0

3 years ago

Why do you not come to thermal equilibrium on a cold day

Tju [1.3M]

Two physical systems are in thermal equilibrium if no heat flows between them when they are connected by a path permeable to heat. Thermal equilibrium obeys the zeroth law of thermodynamics. A system is said to be in thermal equilibrium with itself if the temperature within the system is spatially and temporally uniform.

Systems in thermodynamic equilibrium are always in thermal equilibrium, but the converse is not always true. If the connection between the systems allows transfer of energy as heat but does not allow transfer of matter or transfer of energy as work, the two systems may reach thermal equilibrium without reaching thermodynamic equilibrium.

3 0

3 years ago

What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

3 years ago