How many significant digits are in the measurement 50.003010

True or False: Warm air has greater air pressure than cool air 21 points

alukav5142 [94]

Answer:

False

Explanation:

Warmer air is less dense than cold, which is why warm air tends to rise and cold air sinks. Being acted on by gravity, colder, denser air weighs more and exerts greater pressure per unit area.

8 0

3 years ago

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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

A rope is tied to a box and used to pull the box 2.3 m along a horizontal floor. The rope makes an angle of 30∘ with the horizon

Savatey [412]

Answer:

Answered

Explanation:

A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.

W= FS cosθ

θ= 90 ⇒cos90 = 0 ⇒W= 0

B) work done by tension

W= Tcosθ×S= 5cos30×2.30= 10J

C) Work done by friction force

W= f×s=1×2.30= 2.30 J

D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.

E) The net work done= Work done by tension in the rope - frictional work

=10-2.30= 7.7 J

6 0

3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago

Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te

zlopas [31]

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

4 0

3 years ago