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3 years ago

A particle whose speed is 50 m/sec moves along the line from A(2,1) to B (9,25)

Physics

1 answer:

WINSTONCH [101]3 years ago

3 0

First, calculate for the distance between the given points A and B by using the equation,

D = sqrt ((x2 – x1)2 + (y2 – y1)2)

Substitute the known values:

D = sqrt((9 – 2)2 + (25 – 1)2)

D = 25 m

I assume the unknown here is the time it would require for the particle to move from point A to B. This can be answered by dividing the calculated distance by the speed given above.

t = (25 m)/ (50 m/s) = 0.5 s

Thus, it will take 0.5s for the particle to complete the route.

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For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul

loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

$\frac{P_{o} }{P_{1} } =36.73$ (M₁ = 3)

If Po₁ = Po₂

$\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825$

Table A-1:

$\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63$

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0

3 years ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

The distance the car traveled is 21.45 m

Explanation:

Motion With Constant Acceleration

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

3 years ago

A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is

Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0

3 years ago

Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from

svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2 force of Sun on Mars

F(Merc) = G m M / R^2 force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0

2 years ago

PLEASE HELP ME!!!!!!!

melamori03 [73]

This it you can do it boy or girl 

6 0

3 years ago