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Nady [450]
3 years ago
14

Question 1 of 10

Physics
1 answer:
kirill115 [55]3 years ago
4 0

Answer:

B) trends method

I'm very sure of this answer

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A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Zanzabum

Answer:

Pressure, P = 3724 Pa

Explanation:

Given that,

Depth of water, h_w=20\ cm =0.2\ m

Depth of oil, h_o=40-20=20\ cm=0.2\ m

The density of water, d_w=1000\ kg/m^3

The densinty of oil, d_o=900\ kg/m^3

We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :

P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa

So, the gauge pressure at the bottom of the cylinder is 3724 Pa.

3 0
3 years ago
Please help!<br> Drag and drop the words into the correct locations
Stells [14]

Answer:

background radiation

uniform

13700000000 years ago

7 0
3 years ago
Read 2 more answers
laser light sent through a double slit produces an interference pattern on a screen 3.0 m from the slits. If the eighth order ma
Xelga [282]

Solution :

Given the laser light which is sent through the double slit produces an interference pattern on the screen placed 3  meters from the slits.

The 8th order maximum occurs at angle = 12

So,

$8^{th} \text{ order maxima} = d \sin \theta = m \lambda$      , m = 8

$d = \frac{8 \lambda}{\sin 12}$

$\frac{\lambda}{d}= \frac{\sin 12}{8}$

$3^{rd} \text{ order maxima}= d \sin \theta_2 = m_2 \lambda$

$\sin \theta_2 = \frac{m_2 \lambda}{d}=\frac{3 \lambda}{d}$ $=0.75\ {\sin 12}$

$\theta_2 = \sin^{-1}\left(0.75\ \sin 12\right)$

   $ = \sin^{-1}\left(0.155)$

   $=8.91^\circ$

8 0
3 years ago
Two golfers each hit a ball at the same speed, but one at 60 degrees with the horizontal and the other 30 degrees. a. Which ball
likoan [24]

Answer:

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

Explanation:

Both balls will hit the ground at the same distance and at the same time assuming they both start at the same initial elevation from the ground.

We know that range in the projectile motion given as

R=\dfrac{v_o^2sin2\theta}{g}

sin 60° = sin 120° =0.866

Therefore, the maximum range will be identical assuming both golfers hit the ball at the same elevation from the ground.

As for the time...

Where v is 0 m/s (final velocity at highest point), v0 is the initial velocity (for both golfers) and is the acceleration due to gravity (-9.8 m/s^²). As you can see, the time it takes to get to the highest point is independent of the angle; it is only dependent on the initial velocity. Since both golfers hit the ball with the same speed, the time for the ball to reach the highest point will be the same. Also, since it's a parabola, you can multiply the time by 2 for each golfer to get the time it takes to hit the ground.

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

5 0
4 years ago
A cylindrical tank of methanol has a mass of 70 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific
dem82 [27]

Answer:

Weight=686.7N, \rho=933kg/m^{3}, S.G.=0.933, F=17.5N

Explanation:

So, the first value the problem is asking us for is the weight of methanol. (This is supposing there is a mass of methanol of 70kg inside the tank). We can find this by using the formula:

W=mg

so we can substitute the data the problem provided us with to get:

W=70kg(9.81m/s^{2})

which yields:

W=686.7N

Next, we need to find the density of methanol, which can be found by using the following formula:

\rho=\frac{m}{V}

we know the volume of methanol is 75L, so we can convert that to m^{3} like this:

75L*\frac{0.001m^{3}}{1L}=0.075m^{3}

so we can now use the density formula to find our the methanol's density, so we get:

\rho=\frac{m}{V}

\rho=\frac{70kg}{0.075m^{3}}

\rho=933.33kg/m^{3}

Next, we can us these values to find the specific gravity of methanol by using the formula:

S.G.=\frac{\rho_{sample}}{\rho_{H_{2}O}}

when substituting the known values we get:

S.G.=\frac{933.33kg/m^{3}}{1000kg/m^{3}}

so:

S.G.=0.933

We can now find the force it takes to accelerate this tank linearly at 0.25m/s^{2}

F=ma

F=(70kg)(0.25m/s^{2})

F=17.5N

6 0
3 years ago
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