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3 years ago

3 Points

Physics

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

B.C. D. G.

Explanation:

A vector quantity, has both magnitude and direction. A tip to remember is if you can add a direction to it! You wouldnt say 30 pounds north, but you would say 30 mph north.

<em>I hope this helped! Comment if you have any questions! :)</em>

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Why do planets speed up as they get closer to the sun?:

Serggg [28]

Answer:

Explanation:

Gravity is the main reason that make our planets to pull each other

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2 years ago

While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu

olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

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3 years ago

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

3 years ago

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into t

AnnZ [28]

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, $V_T$ , is given by the sum of the rock, $V_r$ , and the water, V, is given as follows;

$V_T$ = $V_r$ + V

$V_T$ = A × h₂

∴ $V_T$ = 10 cm² × 12 cm = 120 cm³

The total volume, $V_T$ = 120 cm³

The volume of the rock, $V_r$ = $V_T$ - V

∴ $V_r$ = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, $V_r$ = 20 cm³

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

8 0

3 years ago

How to solve 30(a)<br><br> Give solution asap.

expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0

3 years ago