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3 years ago

An object weighs 15N in air and 13N when completely

Physics

1 answer:

anastassius [24]3 years ago

8 0

Answer:

The volume of water displaced = 0.204 m³

Explanation:

From Archimedes' principle,

Upthrust = lost in weight = weight of water displaced.

U = W₁ - W₂..................... Equation 1

Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.

Given: W₁ = 15 N, W₂ = 13 N

Substituting these values into equation 1

U = 15 - 13

U = 2 N

But,

Weight of water displaced = mass of water displaced. × acceleration due to gravity.

Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

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Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

5 0

3 years ago

The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner

Dahasolnce [82]

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

7 0

2 years ago

Please help

ivanzaharov [21]

Answer:

1 +2=3 so thats it

Explanation:

Because i need points

4 0

3 years ago

A. If a T-Rex runs 20 meters in 4 seconds, what is it’s average speed?

mariarad [96]

5m/s

100m

Explanation:

Average speed is sum of distance distance traveled in a given time by a body.

Average speed= $\frac{total distance covered}{time taken}$

Distance = 20m

time = 4s

Average speed = $\frac{20}{4}$ = 5m/s

For the spaceship;

Distance covered = speed x time

Speed = 50m/s

time = 2s

Distance covered = 50 x 2 = 100m

learn more:

Average speed brainly.com/question/5063905

#learnwithBrainly

3 0

3 years ago

A turn signal lamp must be visible in normal sunlight at a distance of at least ______ feet from the front and rear of the vehic

kaheart [24]

Answer:

500; 300 feet.

Explanation:

A turn signal lamp can be defined as an amber blinking lamp which indicates a driver's intent to change direction. It is extremely important that drivers gives a turn signal (flashing light) on the side toward which he or she is turning either left or right.

Simply stated, the turn signal lamp indicate the driver's intent to turn either leftward or rightward by displaying flashing lights to the front and rear of his or her vehicle.

A turn signal lamp must be visible in normal sunlight at a distance of at least 500 feet from the front and rear of the vehicle if the vehicle is at least 80 inches wide, and at least 300 feet from front and rear of the vehicle if the vehicle is less than 80 inches wide according to the transportation traffics code.

8 0

3 years ago