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astraxan [27]
2 years ago
10

8. (Liquids, Gas) have no definite shape, and no definite volume. A)Liquid B)Gas

Physics
1 answer:
Fynjy0 [20]2 years ago
4 0
B. gas has no definite shape or volume. liquid has definite volume but no definite shape
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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
How is the lifetime of a star related to its mass?
Flauer [41]
Stars having less mass collapses early than those with more mass. This can be explained by Einstein's equation E=mc².
According to this equation, mass of stars is converted into light due to thermonuclear reactions occuring in the core of star which acts as engine of the stars. This thermonuclear reactions keeps star alive. Thermonuclear reactions occurs slowly in massive stars hence massive stars live more than light stars.
6 0
3 years ago
A 50-kg copper block initially at 140°c is dropped into an insulated tank that contains 90 l of water at 10°c. Determine the f
xxMikexx [17]

Answer:

T_f=24.71

Explanation:

From the question we are told that:

Mass of block m=50

Temperature of block T_b =140 \textdegree C

Volume of water V= 90L

Temperature of water T_w=10 \textdegree C

Density of water \rho=1000kg/m^3

Specific heat of water C_w=4.18KJ/kg-k

Specific heat of copper C_p=0.96KJ/kg-k

 

Generally the equation for equilibrium stage is mathematically given by

mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)

50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)

48(140-T_f)=376.2(T_f-10)

140-T_f=7.8375(T_f-10)

140-T_f=7.8375T_f-78.375

-8.8375T_f=-218.375

T_f=\frac{-218.375}{-8.8375}

T_f=\frac{-218.375}{-8.8375}

T_f=24.71

6 0
3 years ago
What does the path of an object look like when it is in uniform motion?​
marysya [2.9K]

Answer:

The path of an object in uniform motion is a straight line.

4 0
1 year ago
Read 2 more answers
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