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3 years ago

Help with this. ....

Physics

1 answer:

Dovator [93]3 years ago

4 0

Answer:

Explanation:

The x component is the adjacent side making up the given angle (39.4)

The vector is the hypotenuse.

The definition of the cos (x) is adjacent / hypotenuse.

cos(39.4) = adjacent / 47.3 Multiply both sides by 47.3

47.3 * cos(39.4) = adjacent Cos(39.4) = 0.7727

adjacent = 36.55

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The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of

natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00 m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

$r'=\sqrt{d^2+r^2}$

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

$r'=\sqrt{(1.00)^2+(4.00)^2}$

$r'=\sqrt{1.00+16.00}$

$r'=4.12\ m$

We need to calculate the path difference

Using formula of path difference

$|r'-r|=4.12-4.00$

$|r'-r|=0.12\ m$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Where, v = speed of sound

f = frequency

Put the value into the formula

$\lambda=\dfrac{343}{300}$

$\lambda=1.143\ m$

We need to calculate the phase difference

Using formula of phase difference

$\phi=\dfrac{2\pi\times|r'-r| }{\lambda}$

$\phi=\dfrac{2\pi\times0.12}{1.143}$

$\phi=0.659\ rad$

Hence, The phase difference is 0.659 rad.

7 0

3 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

11111nata11111 [884]

Answer:

(a) $Q = 7.28\times 10^{14}$

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

$\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}$

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

5 0

3 years ago

f the absolute pressure in a tank is 128 kPa , determine the pressure head in mm of mercury. The atmospheric pressure is 100 kPa

White raven [17]

Answer:

211 mmHg

Explanation:

Absolute Pressure = Gauge Pressure + Atmospheric pressure

128 = Gauge Pressure + 100

Gauge Pressure = 28 KPa = 28 × 10³ Pa

Also Gauge Pressure = ρgh

ρ = density = 13550 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = pressure head = ?

28 × 10³ = 13550 × 9.8 × h

h = 28000/(13550 × 9.8)

h = 0.211 m = 211 mm

5 0

3 years ago

A curve of radius 166 m is banked at an angle of 11°. An 736-kg car negotiates the curve at 81 km/h without skidding. Neglect th

lianna [129]

Answer:

$F_n = 7509.33\ N$

Explanation:

given,

radius of curve = 166 m

angle of the banked road = 11°

mass of car = 736 Kg

speed of the curve = 81 km/h

= 81 x 0.278 = 22.52 m/s

normal force acting on the tires

on tire there will be two force acting on it

first one will be force acting due to weight and the other force acting on the tire is due to centripetal force.

$F_n = m g cos \theta+ \dfrac{mv^2}{r} sin \theta$

$F_n = 736 \times 9.8 \times cos 11^0+\dfrac{736 \times 22.52^2}{166} sin 11^0$

$F_n = 7080.28+429.047$

$F_n = 7509.33\ N$

6 0

3 years ago