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julia-pushkina [17]

3 years ago

6

An airplane accelerates from rest down a runway at 9.50 m/s2 for 29.3 seconds when it

1 answer:

Marat540 [252]3 years ago

8 0

Answer:

Explanation:

s = ½at²

s = ½(9.50)(29.3²) = 4,077.8275

s = 4080 m

when rounded to the three significant digits of the question numerals.

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How much current would flow in a device powered by four 1.5 V batteries through a bulb that had a resistance of 51.0 ohms?

Serhud [2]

Answer:

0.12A

Explanation:

Total voltage produced by the batteries;

V = 1.5 × 4 = 6 v

Then the current that is flowing through the device

Current(I) = V/R

; I = 6/51

;Current = 0.12A

4 0

3 years ago

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

As an object falls to the ground its E, is converted to

Andreyy89

Kinetic Energy I’m not 100% shure tho

4 0

3 years ago

Read 2 more answers

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long str

mart [117]

Answer:

$7.3 ms^{-1}$

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

$Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50$

height gained by the ball is given as

$h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m$

$M$ = mass of the ball attached to string = 110 g

$V$ = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

$Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}$

Consider the collision between the two balls

$m$ = mass of the ball fired = 26 g

$v_{o}$ = initial velocity of ball fired before collision = ?

$v_{f}$ = final velocity of ball fired after collision = ?

using conservation of momentum

$m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85$

Using conservation of kinetic energy

$m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}$

3 0

3 years ago