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BARSIC [14]
3 years ago
12

a solid weighs 20gf in air and 18 gf in water.Find the specific gravity of the solid. Please show your work.​

Physics
1 answer:
ikadub [295]3 years ago
5 0

Answer: It is given that A body weighs 20gf in air and 18. 0gf in water. Hence, the answer X-3 = 7.

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A bicyclist travels due east 60.0 km in 3.5 hours. what is the cyclist's average speed?
Reil [10]
I believe it would be 60*3.5.  I apologize if this information is wrong.
5 0
4 years ago
Please help me! Thank you so much!
hodyreva [135]
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4 years ago
When landing after a spectacular somersault, a 40.5 kg gymnast decelerates by pushing straight down on the mat. Calculate the fo
monitta

Answer:

F= = 3.18 * 10^3 N

Explanation:

information given:

m= 40.5 kg

a= seven times gravity so we have

a= 7 g

where

gravity = g  =9.8 m/s2

F= m*a

F= (40.5) * 7g

a = 7 g = (7) (9.8) = 68.6 m/s2

There are two forces acts on gymnsat

Where ,w is the weight of Gymnsat

F(net) = F - mg

F- mg = ma

F = mg+ma

so  

F=  m ( g + a)

F = (40.5) [9.8+ 68.6 ]

F = 3.18 * 10^3 N

4 0
4 years ago
A vector B , with a magnitude of 8.0 m, is added to a vector A , which lies along an x axis. The sum of these two vectors is a t
Paul [167]

Answer:

the magnitude of A is 3.577

Explanation:

According to the question it is given that there is a magnitude of 8.0 m that added to the vector A that lies on x axis also the sum of two vectors would be third vector by considering the y axis and the magnitude is the twice of the magnitude B

The computation of the magnitude of A is shown below:

A \hat{i} +  B  = 2A \hat{j}\\\\A \hat{i} +  8  = 2A \hat{j}\\\\8 = 2A \hat{j} - A \hat{i}\\\\\mid8\mid = \sqrt{(2A \hat{j})^2 + (- A \hat{i})^2} \\\\8 = \sqrt{5A} \\\\A = \frac{8}{\sqrt{5} }

A = 3.577

Hence, the magnitude of A is 3.577

7 0
3 years ago
A monatomic ideal gas that is initially at a pressure of 1.50 * 105 Pa and has a volume of 0.08 m3 is compressed adiabatically t
Katyanochek1 [597]

Answer:

1.571 is the required ratio

Solution:

As per the question:

Pressure, P = 1.50\times 10^{5}\ Pa

Volume, V = 0.08\ m^{3}

Volume,V' = 0.04\ m^{3}

(a) To calculate the final pressure:

In adiabatic compression:

PV^{\gamma} = P'V'^{\gamma}

\gamma = 1.67

P' = (\frac{0.04}{0.08})^{1.67}\times 1.50\times 10^{5}

P' = 4.713\times 10^{5}\ Pa

(b) From the eqn of ideal gas:

PV = nRT

We can write:

\frac{P'V'}{PV} = \frac{T'}{T}

where

T' = Final temperature

T = Initial temperature

Thus

\frac{4.713\times 10^{5}\times 0.04}{1.50\times 10^{5}\times 0.08} = \frac{T'}{T}

\frac{T'}{T} = 1.571

7 0
3 years ago
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