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3 years ago

10

When a fish is floating motionless below the surface of a lake, what is the amount and direction of the force the water is exert

ing on it?

1 answer:

miskamm [114]3 years ago

3 0

Answer:

Buoyant force which is equal in magnitude to the fish's weight

Explanation:

A fish floating motionless below the surface of a lake, is an equilibrium position with the water.

Hence, upward force must be equal to downward force.

The downward force is due to fish's weight while the buoyant force is the upward force due to force of the water.

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A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm

Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

$t=5\cdot 10^{-5}s$

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

$c=3.0\cdot 10^8 m/s$

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

$d=vt$ (1)

where

$v=c=3.0\cdot 10^8 m/s$

$t=5\cdot 10^{-5}s$

$d=2L$ , where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

$L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m$

7 0

4 years ago

Hurryyyyy plzzzzzz..........the one in the middle...

zheka24 [161]

Answer:

Friction force is independent of the direction of the contacting surfaces

Explanation:

It can go any way depending on how much force is being out on it.

5 0

3 years ago

An Argon laser (λ = 5.0×102nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the speed of

Yanka [14]

Answer:

The speed of the laser light in the cable, $c_f=2.1\times 10^8\ m/s$

Explanation:

It is given that,

Wavelength of Argon laser, $\lambda=5\times 10^2\ nm=5\times 10^{-7}\ m$

Refractive index, n = 1.46

Let $c_f$ is the speed of the laser light in the cable. The speed of light in a medium is given by :

$c_f=\dfrac{c}{n}$

$c_f=\dfrac{3\times 10^8\ m/s}{1.46}$

$c_f=2.05\times 10^8\ m/s$

or

$c_f=2.1\times 10^8\ m/s$

So, the speed of the laser light is $2.1\times 10^8\ m/s$ . Hence, this is the required solution.

4 0

4 years ago

Read 2 more answers

Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

8 0

4 years ago

A seventh grade class is studying the solar system. A small group of students wanted to expand their studies to include the form

rodikova [14]

Answer:by. III

Explanation:

7 0

3 years ago

Read 2 more answers