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3 years ago

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a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft

, what is the pressure in lb/ft2 at the bottom of the tank

1 answer:

vlabodo [156]3 years ago

4 0

Answer:

<u><em>note:</em></u>

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

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What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s

stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0

2 years ago

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

tamaranim1 [39]

We have that the amount of the charge q is

$q=1.8*10^{-7}$

From the Question we are told that

Force $F=7.89 x*10^{-7}$

Velocity $V=2090m/s$

Angle $\theta=29.4$

Magnetic field $B=4.23 * 10^{-3} T$

Generally, the equation for Force F is mathematically given by

$F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}$

$q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}$

$q=1.8*10^{-7}$

In conclusion

The amount of the charge q is

$q=1.8*10^{-7}$

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brainly.com/question/1470439

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2 years ago

Read 2 more answers

A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0

2 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Vsevolod [243]

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

3 0

3 years ago

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = <em>0.97c</em>

3 0

2 years ago