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Angelina_Jolie [31]

3 years ago

12

PLEASE HELP!

2 answers:

tino4ka555 [31]3 years ago

8 0

Answer:

option (c)

Explanation:

A small diameter open tube is called capillary tube.

When a capillary tube is immersed in a vessel containing mercury, then the mercury falls in the capillary tube and the shape is like outside of an umbrella.

As the angle of contact for mercury and glass is obtuse, so it falls in the tube and the meniscus is convex.

maks197457 [2]3 years ago

4 0

The answer is A .
Due to air pressure, the mercury in the tube will be higher than that in the vessel.
Mercury in a container, the center of the liquid is always higher, so it is shaped like the outside of an umbrella.

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Kisachek [45]

Answer:

All points to the left of zero are negative

Explanation:

6 0

2 years ago

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Salmon often jump waterfalls to reach their

PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0

2 years ago

A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?

Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

$\alpha =\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

$\omega=25 rev/min$

This means that the change in angular velocity is zero:

$\Delta \omega=0$

And so, that the angular acceleration is zero:

$\alpha=0$

8 0

3 years ago

When a bat hits a ball, the impulse is what?

german

I believe the answer is C

4 0

3 years ago

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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

3 years ago