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Lerok [7]
1 year ago
5

A solid weighs 0.9N in air and 0.2N in a liquid of density 700kg/m^3. Calculate

Physics
1 answer:
Setler [38]1 year ago
7 0

The correct answer is Upthrust= 0.7N and Volume = 1.02 x 10⁻² g

Solid weighs.09 N in air.

Solid weighs less than liquid because of an upward force operating on it (solid weighs in liquid =.02 N).

A solid body experiences an upthrust when it is partially or entirely submerged in a fluid, such as water. In this case, the fluid's weight (water) equals the fluid's weight that the water has displaced.

upthrust is = to.09 -.02 N = 0.7N.

Upthrust = displaced liquid's weight

so that weight of liquid equal to volume of solid = 0.7 N

Volume of solid times mass of liquid = 0.07/9.8 = 0.00714 =  7.14 x 10⁻³ kg

Volume = Mass / density

Volume = 7.14 x 10⁻³ / 700

= 1.02 x 10⁻⁵ kg= 1.02 x 10⁻² g

To learn more about upthrust refer the link:

brainly.com/question/24383048

#SPJ9

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4 0
4 years ago
What is your question?
Papessa [141]
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150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit
zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

I=\frac{q}{t}

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For the cell in this problem, we have

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I=\frac{150}{30}=5 A

Learn more about current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

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6 0
3 years ago
Why can't there be a number lower than absolute zero
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5 0
3 years ago
Read 2 more answers
In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

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4 0
3 years ago
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