If a cart of 4 kg mass has a force of 8 newtons exerted on it, what is its acceleration

Mechanical (sound) waves to Earth from satellites. How is this possible? aves are unable to travel through a vacuum, such as thr

Anuta_ua [19.1K]

Sound waves are not able to travel in a vacuum, sound requires a medium such as air or a solid for example. Satellites are indeed in space, but radio waves are not sound waves they are a form of light. Light can travel in a vacuum so the messages that satellites beam back to Earth are light waves not sound waves, that is why it is possible.

6 0

2 years ago

A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant

Likurg_2 [28]

Skskdidododododoorrororo

3 0

2 years ago

in the design a thermos lab, you compared the temperature of your thermos with a container that you did not insulate. what was t

irina [24]

To see if the insulation would affect the temp of whatever you are measuring. <span />

4 0

2 years ago

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

2 years ago

How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th

Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

$R_t=\dfrac{r_1\times r_2}{r_1+r_2}$

Taking reciprocal of the above equation.

$\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}$

We can also write the above equation as :

$\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}$

On simplification of above equation,

$\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}$

or

$\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}$

Hence, proved.

4 0

2 years ago