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4 years ago

If racing alcohol has a mass density of 790kg/m3, what mass will a 1250-litre tank hold?

1 answer:

3 0

Density = Mass per unit Volume
that is,
D = m / V

Now,
We've been provided with,

Density = 790 kg/m³
Volume = 1250 litres = 1.250 m³

Now,
$density = \frac{mass}{volume} \\ 790 = \frac{m}{1.250} \\ 790 \times 1.250 = m \\ m = 790 \times 1.250 \\ m = 790 \times \frac{1250}{1000} \\ m = 790 \times \frac{5}{4} \\ m = 197.5 \times 5 \\ m = \frac{1975}{10} \times 5 \\ m = \frac{1975}{2} = 987.5 \: kg$

987.5 kg of mass will a 1250 litres tank will hold.

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A researcher measures the thickness of a layer of benzene (nn = 1.50) floating on water by shining monochromatic light onto the

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Answer:

The minimum thickness is $t= 8.75*10^{-8} m$

Explanation:

generally the equation for thin film interference is mathematically represented as

$2nt = (m + \frac{1}{2} ) \lambda$

Where t the thickness

m is any integer

n is the refractive index of the film

$\lambda$ is the wavelength of light

Since we are looking for the thickness we make t the subject of the formula

$t = \frac{(m+ \frac{1}{2} ) \lambda}{2n}$

m= 0 cause the thickness is minimum at m=0

Substituting values

$t = \frac{(0 +\frac{1}{2}) 8525*10^{-9} }{2 *1.5}$

$t= 8.75*10^{-8} m$

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3 years ago

Read 2 more answers

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave

Marina86 [1]

Answer:

$a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree$

Explanation:

Puck A's initial speed is $v_a_1=14.6m/s$ and move in a direction $\theta_b=28.0\textdegree$ after the collision.

# $P_1=P_2$ since there's no external force on the system( $P=mv).$

#The collision equation can be written as;

$m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2$

The kinetic energies before and after the collision are expressed as:

$K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2$

$0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2$

Let +x along A's initial direction and +y along A's final direction makes the angle $62\textdegree$

$\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j$

#Substitute in $v_{b2}^2=213.16-v_{a2}^2$ :

$\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\$

$v_{a2}=12.891\ m/s$

Hence, the the speed of puck A after the collision is 12.891 m/s

#. The velocity of A after the collision is;

$\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j$

Substitute $\dot v_{a2}$ into $\dot v_{b2}=(14.6i)-\dot v{a2}$ :

$\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j$

This is the velocity of puck B after the collision, it's speed is:

$v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s$

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

$\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree$

Hence, the direction of B's velocity after the collision is 62°

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4 years ago

Summary about down hill on a bicycle story

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Answer:

The poem, Going Down the Hill on a Bicycle, written by Henry Charles Beeching describes the thrilling ride of a boy going downhill. ... The poet mentions how he lifts his feet from the pedals and keeps his hands still so that he would not lose his balance and fall off the bicycle, while it is dashing down the hill.

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3 years ago

Energy balance worksheet

m_a_m_a [10]

I think it’s current= A / C

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3 years ago

A baseball bat contacts a 0.145-kg baseball for 1.3×10−3 s The average force exerted by the bat on the ball is 8900 N. If the ba

Nezavi [6.7K]

Answer: v2 = -50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat. (Reverse direction)

Explanation:

Using the law of conservation of momentum:

The impulse of the bat = change in momentum of the baseball

Ft = ∆M

Ft = m1v1 - m2v2

Since m1 = m2 = m (the mass remains unchanged)

Ft = m(v1 - v2) .....1

Given:

Force F = 8900N

Mass m = 0.145kg

time t = 1.3 × 10^-3 s

Initial velocity v1 = 29m/s

Substituting into eqn 1

8900(1.3×10^-3) = 0.145(29-v2)

29-v2 = 79.79

v2 = 29 - 79.79 = -50.8m/s

v2 = -50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat.

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4 years ago