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Serhud [2]
2 years ago
11

An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object

Physics
1 answer:
Elden [556K]2 years ago
8 0

Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2  = 1.5  N  of buoyancy

density of water = 1 gm /cc

 1.5 N = m (9.81)

    m of water displaced = .1529 kg

             152.9 cc of water will produce this buoyancy....this is the volume of the object

   

find mass of object     2.7 = m (9.81)     shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

  ====> it is 1.8 times denser than water and will sink when in water....

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A boy on the beach holds a spherical balloon filled with air. at 10:00am, the temperature on the beach is 20°c and the balloon h
Anon25 [30]
For idea gases, volume is directly proportional to temperature. That is, an increase in temperature leads to increase in volume and vice versa.

Therefore,
V1/T1 = V2/T2 => T2 = (V2*T1)/V1

Assuming that the balloon is spherical in shape,

V= 4/3*pi*R^3.... In the formula for calculating T2, 4/3*pi cancels out.

R1 = 30/2 15 cm; R2 = 30.5/2 = 15.25 cm; T1 = 20+273.15 =293.15 K

Therefore,

T2 = (R2^3*T1)/R1^3 = (15.25^3*293.15)/15^3 = 308.05 K = 34.9 °C
6 0
3 years ago
Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

3 0
3 years ago
Einstein’s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had
PtichkaEL [24]

Answer:

1. bending of light in gravitational fields.

2. effect of gravitational redshift.

3. perihelion precission of mecury.

Explanation:

1 bending of light in gravitational fields, we can think of it like this:

by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.

2. effects of gravitational redshift:

this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.

3. perihelion precission of mecury:

according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.

3 0
4 years ago
You have a mass of 95 kg.<br> a. What is your weight on Earth?
svp [43]

Answer:

931.63N

Hope this helps :D

8 0
3 years ago
A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d
prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

V_{t}= V_{o}e^{\frac{-t}{R*C} }

Voltage in this case is the energy dissipated so

E_{t}= E_{o}e^{\frac{-t}{R*C} }

\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }

\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

Using the equation to find capacitance

ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

8 0
3 years ago
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