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DIA [1.3K]
3 years ago
13

Help? 25 points and will give brainliest!

Physics
1 answer:
ladessa [460]3 years ago
8 0

Answer: 44.89 times

Explanation:

Given that the earth's average velocity is about 6.7 times faster than that of planetoid Makemake

Given that the formula involved is:

R2/R1 = (V1/V2)^2

Let V1 = earth velocity and R1 = earth orbital distance

While R2 = Makemake velocity and R2 = orbital distance.

Since the earth's average velocity is about 6.7 times faster than that of planetoid Makemake, then,

V1/V2 = 6.7

Substitute this in the formula above

R2/R1 = (6.7)^2

R2/R1 = 44.89

Cross multiply

R2 = 44.89R1

Therefore, Makemake's orbit is about 44.89 times larger than the earth's orbit.

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Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
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where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
\tan \alpha  =  \frac{3}{10} =0.3
And so the angle is
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Which tool would you use to measure the amount of rainfall?
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A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
In a certain electrolysis experiment involving Al3+ ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many
yawa3891 [41]

Answer:

30.5 x 10³ min.

Explanation:

Atomic weight of aluminium = 27

For the reaction

Al⁺³ + 3e  = Al

3 mole of electron is required by 1 mole of aluminium

3 x 96500 C of charge is required for 27 gram of aluminium

60.2 g aluminium will require

\frac{3\times96500}{27} \times60.2

645.47 x 10³ C

So charge required = 645.47 x 10³ C

Now Charge  = Current x time

Current ( given ) = 0.352A

Time = charge / current = 645.47 x 10³ / .352

=1833 x 10³ s

30.5 x 10³ minutes

6 0
3 years ago
6.7.1 Current and Current Density A sphere of radius 10 mm that carries a charge of 8 nC =8x 10°C is whirled in a circle at the
aev [14]

Answer:

Part a)

Rate of charge flow is known as electric current

Part b)

Average current flow is

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

j = 1.27 \times 10^{-3} A/m^2

Explanation:

As we know that angular frequency of rotation is

\omega = 100\pi rad/s

now by basic definition of electric current

Part a)

Rate of charge flow is known as electric current

i = \frac{dq}{dt}

so here we have

i = \frac{Q}{T}

i = Qf

Part b)

here we know that

\omega = 2\pi f

100\pi = 2\pi f

f = 50 Hz

now we have

i = (8\times 10^{-9})(50)

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

So we have

j = \frac{i}{A}

j = \frac{4\times 10^{-7}}{\pi(10\times 10^{-3})^2}

j = 1.27 \times 10^{-3} A/m^2

6 0
3 years ago
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