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3 years ago

Help solve these two problems im having trouble trying to start these problems?

Physics

1 answer:

belka [17]3 years ago

6 0

Answer:

25. Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

$d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m$

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

$east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km$

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

$north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km$

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A particle with a charge of 2e moves between two points which have a potential difference of 75V. What is the change in potentia

Sonbull [250]

Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.

In our case:

Q=2e=2*(-1.6*10^-19) C
V=75 V

Ep=(-3.2*10^-19)*75

Ep=-2.4*10^-17 J

The change in potential energy of the charge is -2.4*10^-17 J

5 0

3 years ago

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretch

r-ruslan [8.4K]

Answer:

a) A = 0.98 m

b) Ф = 90°

c) x = -0.98sin(12.25t)

Explanation:

We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s

The expression for the amplitude under these conditions is:

A = √xo² + vx²/w² (1)

To calculate the angular speed w, we use the following expression:

w = √k/m (2)

Calculating w:

w = √300/2 = 12.25 rad/s

Now, we replace this value into equation 1, along with the other known values and solve for A:

A = √0 + (12)²/(12.25)²

A = 0.98 m

b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

xo/A = cosФ

Ф = arccos(x0/A) (6)

Replacing the data in (6):

Ф = arccos(0/0.98)

Ф = 90°

c) Equation (4) is the expression for the simple harmonic motion

x = A cos(wt - Ф)

And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

And we have an trigonometric expression for cos that is:

cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

cos(α - π/2) = cos(12.25t - 90)

x = -0.98sin(12.25t)

3 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago

ALL OF MY POINTS FOR THIS!

scoray [572]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

   = (1,127 kg) x (6 m/s² forward)

   = (1,127 x 6) newtons forward

   = 6,762 newtons forward
______________________________

   Momentum = (mass) x (speed)

   =   (69 kg) x (6 m/s)

   = 414 kg-m/s

3 0

3 years ago

You have to run 2.2 miles in track. How far is this in feet? Note: There are

Romashka [77]

Answer: B

Explanation: This can be easily done by inputting 5280 * 2.2.

3 0

3 years ago

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Help solve these two problems im having trouble trying to start these problems?​

Help solve these two problems im having trouble trying to start these problems?