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Rina8888 [55]
3 years ago
13

Which of the following statements about Masters programs is not correct?

Physics
1 answer:
uranmaximum [27]3 years ago
7 0

The correct answer is C. The level of competition is not very high in most Masters  programs.

Explanation:

In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters  programs".

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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
QveST [7]

Answer:

Please see attachment

Explanation:

Please see attachment

8 0
3 years ago
) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick
sveticcg [70]

Answer:

  about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

  y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

  -13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

  tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

  α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

_____

I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

5 0
2 years ago
Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) Wha
k0ka [10]

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

3 0
3 years ago
A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?​
ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut    <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0
3 years ago
A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
3 years ago
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