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Ray Of Light [21]

3 years ago

15

Which of the following terms refers to any push or pull on an object

1 answer:

Murljashka [212]3 years ago

3 0

Its either B. force or A. net force but I will go with force because forces can either push or pull I hope it help you.

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How can you use a magnet to induce a current?

Furkat [3]

A coil of wire with a current flowing thru it becomes a magnet

3 0

3 years ago

Read 2 more answers

A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

6 0

3 years ago

calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s

zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0

2 years ago

What are the unit for acceleration

Flura [38]

<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

----------------

time

= m/s

------

s

=m/s^2

7 0

3 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago