A 23.8 kg space rock is pushed with a force and accelerates at 8.97 m/s/s. How much force was applied?

An electric motor rotates 60 times per second if the alternating current source is 60 Hz. How many times will an electric motor

valentina_108 [34]

Answer:

180,000

Explanation:

Frequency is a quantity that is measured in Hertz [Hz] and it represents the number of rotations per second.

A motor with a frequency of 50 Hz will rotate 50 times per second.

Since we don't want to know how many times it rotates per second, but per hour. The first step is to find how many seconds there are in an hour and then multiply that amount by 50.

Seconds in an hour:

there are 60 seconds per minute, and 60 minutes per hour, thus there are

60*60 = <u>3,600 seconds in an hour</u>

We know that the motor will rotate 50 times per second so to find the number of rotations in 1 hour = 3,600 seconds we multiply:

50*3,600 = 180,000 rotations

8 0

3 years ago

PLSSS HELP MEEEE

vekshin1

Answer:

Cu2+

Explanation:

7 0

3 years ago

Read 2 more answers

A car enters a horizontal, curved roadbed of radius 50 m. the coefficient of static friction between the tires and the roadbed i

UkoKoshka [18]

Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
v = √(kgr)
v = √(0.20·9.8 m/s²·50 m)
= 7√2 m/s ≈ 9.90 m/s

6 0

3 years ago

Read 2 more answers

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

4 years ago

Elsie is finishing second grade. If she goes to school 147 day per year and she have 10 years of school left, how many days of s

ICE Princess25 [194]

Answer:

1,323 days left

Explanation:

147 x 10 = 1,470

1470 - 147 = 1,323

Hopefully this helps you :)

pls mark brainlest ;)

6 0

3 years ago