What does Einstein's famous equation for nuclear energy, E = mc^2, mean?

Question 27 a motorboat takes 3 hours to travel 144km going upstream. the return trip takes 2 hours going downstream. what is th

Vedmedyk [2.9K]

In this item, we let x be the rate of the boat in still water and y be the rate of the current.

Upstream. When the boat is going upstream, the speed in still water is deducted by the speed of the current because the boat goes against the water. The distance covered is calculated by multiplying the number of hours and the speed.
(x - y)(3) = 144

Downstream. The speed of the boat going downstream is equal to x + y because the boat goes with the current.
(x + y)(2) = 144

The system of linear equations we can use to solve for x is,
3x - 3y = 144
2x + 2y = 144

We use either elimination or substitution.

We solve for the y of the first equation in terms of x,
y = -(144 - 3x)/3

Substitute this to the second equation,
2x + 2(-1)(144 - 3x)/3 = 144
The value of x from the equation is 60

<em>ANSWER: 60 km/h</em>

5 0

3 years ago

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame

natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

$\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}$

where;

L = length of the bed

$\mu$ = viscosity

U = superficial velocity

$\epsilon$ = void fraction

dp = equivalent spherical diameter of bed material (m)

$\rho$ = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6 -------------- equation (1)

24A + 576B = 24.1 --------------- equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

B = 0.017014

From equation (1)

12A + 144B = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

$A = \frac{9.6 -144(0.017014}{12}$

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0

3 years ago

A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *

ad-work [718]

Answer:

D) 15 m/s

Explanation:

v = u + at

v = 10 m/s + 1 m/s²(5 s)

v = 15 m/s

3 0

2 years ago

Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.

Margarita [4]

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light = ?

Solution:

The wavelength can be derived using the expression below;

wavelength = $\frac{v}{f}$

v is the speed of light

f is the frequency

wavelength = $\frac{3 x 10^{8} }{5.085 x 10^{14} }$ = 5.9 x 10⁻⁷m

3 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$