A remote-controlled car’s wheel accelerates at 22.7 rad/s2 . If the wheel begins with an angular speed of 10.3 rad/s, what is the wheel’s angular speed after exactly twenty full turns

Answers

Answer 1

Answer:

Explanation:

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Where is information first stored in a human brain

Answers

Answer:

when u learn something new it goes to ur short term memory

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = $P= (RT)/(√(2)N_A\sigma\lambda )$

putting values we get

= $(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )$

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

$Electric FIeld = (Change\: in\: Potential)/(Distance)$

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance= $0.8 * 10^(-2) m$

So, Electric field is

$Electric\: field=(3.8V)/(0.8 * 10^(-2)m )$

$Electric\: field=475 V/m$

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

A 60kg woman on skates throws a 3.9kg ball with a velocity of37m.s west. What is the velocity of the woman?

Answers

Answer:

2.405 m/s

Explanation:

Given that,

Mass of a women, m₁ = 60 kg

Mass of a ball, m₂ = 3.9 kg

Velocity of the ball, v₂ = 37 m/s

We need to find the velocity of the woman. It is a concept based on the conservation of linear momentum. Let v₁ is the velocity of the woman. So,

$m_1v_1=m_2v_2\n\nv_1=(m_2v_2)/(m_1)\n\nv_1=(3.9* 37)/(60)\n\nv_1=2.405\ m/s$

So, the velocity of the woman is 2.405 m/s.

A person is making homemade ice cream. She exerts a force of magnitude 26 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.26 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 2.0 s, what is the average power being expended?

Answers

Answer:

P = 31.83 W

Explanation:

Our data are,

Magnitude of the force F = 26 N

Radius of the circular path r = 0.26 m

The angle between force and handle $\theta = 0$ °

Time t = 2 s

We know that the formula to find the velocity is given by

Velocity $v = (2\pi r)/(t)$

$v= (2\pi r)/(t)$

$v=(2 \pi 0.26)/(2)$

$v= 0.8168m/s$

We know also that the formula to find the power is given by,

$P = F*v$

$P = (26)(0.8168)$

$P = 31.83 W$

An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.Decrease the mass of the car slightly.
Increase the mass of the car slightly.
Decrease the launch speed of the car slightly.
Increase the launch speed of the car slightly.

Answers

Answer:

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

Explanation:

Kinetic energy is

K = ½ m v²

the speed of the expensive we can find it r

v² = v₀² + 2 a x

we can find acceleration with Newton's second law

F = m a

a = F / m

F= cte

substitute in the velocity equation

v² = v₀² + 2 F/m x

let's substitute in the kinetic energize equation

K = ½ m (v₀² + 2 F/m x)

K = ½ m v₀² + f x

we see that the kinetic energy depends on two tomines

in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy

As the launch speed increases the initial energy increases quadratically

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

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