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3 years ago

In which direction does gravitational force act

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

4 0

Gravitational force pushes down on us to keep us on the ground. So it pushes in a downward direction.

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if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate

777dan777 [17]

The sound power the person generated is $2.313 \times 10^{-5} \ W$ .

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

$A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2$

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

$P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W$

Learn more about intensity of sound here: brainly.com/question/4431819

4 0

2 years ago

The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b

ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})$

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) = \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V$

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

8 0

3 years ago

Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t

Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0

3 years ago

A sharp raises a note by

vivado [14]

The answer should be B. A half step

8 0

3 years ago

Read 2 more answers

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

Replacing v₀, at and t in (1), we have:

$x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

Therefore, as the truck travels twice as far as the car, the right answer is a).

7 0

3 years ago