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3 years ago

The contact force that acts on objects in a liquid or gas and allows objects to float is called

Physics

2 answers:

Harrizon [31]3 years ago

8 0

<span>The contact force that acts on objects in a liquid or gas and allows objects to float is called </span>Buoyancy.

bogdanovich [222]3 years ago

7 0

the answer is Buoyancy

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Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes,

zheka24 [161]

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta P = \dfrac{128 \mu L Q}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D = D_A = D_B$

where;

length of the first pipe A $L_A = L$ and the length of the second pipe B $L_B = 2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}$

$\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}$

$\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}$

$\mathbf{Q_1 = 2Q}$

6 0

3 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

When using the magnification equation, a value greater than 1 as the solution for M indicates that the image ? is larger than th

Nadusha1986 [10]

M= Height of image/height of the object

If, M>1, then height of image>height of object.

And therefore, image is larger than the object.

7 0

4 years ago

Read 2 more answers

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

3 years ago

what term describes the situation where the frequency of the wind match the frequency of the building? a. velocity b. resonance

saul85 [17]

The answer is B) resonance

3 0

3 years ago