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Zanzabum
2 years ago
12

A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large

as the droplet?
Physics
2 answers:
luda_lava [24]2 years ago
6 0
A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large as the droplet?

Answer: In this case the density will remain constant for both droplets. The reason being that volume will not change the density of the material. The only way of changing it is by changing its state. If you increase the volume then the mass will also increase. Leaving the density the same.

I hope it helps, Regards.
Llana [10]2 years ago
6 0
I think it would be the same

density is an intensive property of the matter, it doesn't change with the change of mass for the same substance, at the same pressure and temperature

hope this helps
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Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative
Evgesh-ka [11]

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC /  AB

we substitute

tan( ∠ ABC ) = 7 /  45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC )  is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

4 0
2 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches t
kramer

Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

4 0
2 years ago
A transverse wave on a string has an amplitude a. A tiny spot on the string is colored red. As one cycle of the wave passes by
aliya0001 [1]

Answer:

Option D) 4A

Explanation:

As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .

Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e

* Upward movement through distance A

*Downward movement through distance A

*Downward again through distance A

*Upward through distance A.

Then it would travel back to its undistributed position held

4 0
2 years ago
I'm not really sure how to go about creating the equation, can anyone help me?
AlexFokin [52]
The displacement vector (SI units) is
\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}

The speed is a scalar quantity. Its magnitude is
v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}

Answer: At√(t⁴ - 12t³ + 36t² + 1)
3 0
3 years ago
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