Ramon and Sally are observing a toy car speed up as it goes around a circular track. Ramon says, “The car’s speeding up, so there must be a net force parallel to the track.” “I don’t think so,” replies Sally. “It’s moving in a circle, and that requires centripetal acceleration. The net force has to point to the center of the circle.” Do you agree with Ramon, Sally, or neither

The planet's orbital period is about 388 days

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of the planet = m = 6.75 × 10²⁴ kg

mass of the star = M = 2.75 × 10²⁹ kg

radius of the orbit = R = 8.05 × 10⁷ km = 8.05 × 10¹⁰ m

Unknown:

Orbital Period of planet = T = ?

Solution:

Firstly , we will use this following formula to find the orbital period:

[tex]F = ma[/tex]

[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]

[tex]G M = \omega^2 R^3[/tex]

[tex]\frac{GM}{R^3} = \omega^2[/tex]

[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]

[tex]T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}[/tex]

[tex]T \approx 3.35 \times 10^7 \texttt{ seconds}[/tex]

[tex]T \approx 388 \texttt{ days}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

To calculate the orbital period of the planet, convert the radius to meters, substitute the mass of the star and the radius into Kepler's third law, solve for the orbital period squared, and convert to Earth days.

The question involves applying Kepler's third law of planetary motion to calculate the orbital period of a planet revolving around a star. The law relates the orbital period (T) to the radius (r) of the orbit and the mass (M) of the star around which the planet orbits. Kepler's third law can be expressed as:

T² = (4π²/GM) · r³,

where G is the gravitational constant (6.67430 × 10⁻¹¹ m³/kg · s²).

To find the orbital period of the planet:

The student will then have the orbital period of the planet in Earth days.

Let [tex]\theta[/tex] be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

[tex]\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

The current has velocity vector (relative to the Earth)

[tex]\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath[/tex]

The swimmer's resultant velocity (her velocity relative to the Earth) is then

[tex]\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}[/tex]

[tex]\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath[/tex]

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

[tex]\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ[/tex]

which is approximately 41º west of north.

The swimmer should swim approximately 36.87 degrees in the north-west direction to counteract the eastward ocean current and reach Victoria, British Columbia directly from Port Angeles, Washington. This is derived by considering the velocity vectors of the swimmer and the current.

This question is about the interaction of velocity vectors which is a concept from physics. The swimmer's speed and the ocean current each represent a vector with both a magnitude (speed) and direction. We must factor in these two vectors to work out where the swimmer should aim to swim.

Assuming the north direction as +y axis and east as +x axis, the swimmer intends to swim north with velocity 4.0 km/h and the water's current is moving towards east with velocity 3.0 km/h. In order for the swimmer to move directly north, the swimmer needs to swim against the current, which means westwards. By utilizing the Pythagorean Theorem and trigonometric principles, we can calculate the angle of her direction which is arctan(3/4) in the north-west direction or around 36.87 degrees from north.

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[tex]\boxed{P=96.09MW}[/tex]

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

[tex]R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega[/tex]

So we can calculate the power loss as:

[tex]P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}[/tex]

Finally, the power loss due to resistance in the line is 96.09MW

Answer:

4582 N

Explanation:

The initial speed of the test car is

u = 11.4 m/s

While the final speed is

v = 0

The displacement of the test car during the collision is

d = 0.78 m

So we can find the acceleration of the car by using the following SUVAT equation:

[tex]v^2 - u^2 = 2ad\\a=\frac{v^2-u^2}{2d}=\frac{0-(11.4)^2}{2(0.78)}=-83.3 m/s^2[/tex]

Now we can find the average force acting on the dummy by using Newton's second law:

F = ma

Where m = 55 kg is the mass. Substituting,

[tex]F=(55 kg)(-83.3 m/s^2)=-4582 N[/tex]

So the size of the average force is 4582 N.

Answer:

167.85 rev / s

Explanation:

r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N

F = m r w^2

where, w is the angular velocity.

4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2

w = 1054.1 rad / s

w = 2 π f

f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s

The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.

Recall :

We calculate the angular velocity, ω thus :

ω² = F/mr

ω² = [tex] \frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}[/tex]

ω = [tex] \sqrt{1.11 \times 10^{6}} = 1053.56 rad/s[/tex]

Frequency = 1053.56 ÷ (2π)

Frequency = 167.68 Hz

Therefore, the Number of revolutions per seconds would be about 167.8 Hz

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Answer:

82 pF

Explanation:

diameter = 8.2 cm, distance, d = 2.1 mm = 0.0021 m,

dielectric constant, k = 3.7

radius = 1/2 x diameter = 8.2 / 2 = 4.1 cm = 0.041 m

The formula for the capacitance of parallel plate capacitor is given by

[tex]C = \frac{k \varepsilon _{0}\times A}{d}[/tex]

[tex]C = \frac{3.7 \times 8.854\times 10^{-12} \times \pi \times 0.041 \times 0.041}{0.0021}[/tex]

C = 8.23 x 10 ^-11 F

C = 82 pF

(a) 1) The field is in the direction of the electron's initial velocity

The electric field is in a direction opposite to the initial velocity of the electron.

Let's remind that, when an electric charge is immersed in an electric field:

- if the charge is positive, the charge experiences a force in the same direction as the electric field direction

- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction

In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.

(b) [tex]4.13\cdot 10^{-4} m[/tex]

We have:

Electron's initial velocity: [tex]u=5.25\cdot 10^6 m/s[/tex]

Electric field magnitude: [tex]E=1.9 \cdot 10^5 N/C[/tex]

Electron charge: [tex]q=-1.6\cdot 10^{-19} C[/tex]

Mass of the electron: [tex]m=9.11\cdot 10^{-31}kg[/tex]

The electric force exerted on the electron is:

[tex]F=qE=(-1.6\cdot 10^{-19} C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^{-14}N[/tex] (the negative sign means the direction of the force is opposite to its initial velocity)

The electron's acceleration is given by:

[tex]a=\frac{F}{m}=\frac{3.04\cdot 10^{-14} N}{9.11\cdot 10^{-31} kg}=-3.34\cdot 10^{16} m/s^2[/tex]

Now we can use the SUVAT equation:

[tex]v^2 - u^2 = 2ad[/tex]

where

v = 0 is the final speed (the electron comes to rest)

d is the total distance travelled by the electron

Solving for d,

[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(5.25\cdot 10^6 m/s)^2}{2(-3.34\cdot 10^{16} m/s^2)}=4.13\cdot 10^{-4} m[/tex]

(c) [tex]1.57\cdot 10^{-10}s[/tex]

We can use the following equation:

[tex]a=\frac{v-u}{t}[/tex]

where we have

[tex]a=-3.34\cdot 10^{16}m/s^2[/tex] is the electron's acceleration

v = 0 is its final speed

[tex]u=5.25\cdot 10^6 m/s[/tex] is the initial speed

t is the time it takes for the electron to come at rest

Solving for t,

[tex]t=\frac{v-u}{a}=\frac{0-(5.25\cdot 10^6 m/s)}{-3.34\cdot 10^{16} m/s^2}=1.57\cdot 10^{-10}s[/tex]

(d) [tex]5.25\cdot 10^6 m/s[/tex]

This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.

So we have:

u = 0 is the initial speed of the electron

[tex]a=3.34\cdot 10^{16}m/s^2[/tex]

[tex]d=4.13\cdot 10^{-4} m[/tex] is the distance covered to go back

So we can use the following equation:

[tex]v^2 - u^2 = 2ad[/tex]

to find v, the new final speed:

[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2 + 2(3.34\cdot 10^{16} m/s^2)(4.13\cdot 10^{-4} m)}=5.25\cdot 10^6 m/s[/tex]

Answer:

17 seconds

Explanation:

In the y direction:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 140 + (120 sin 38) t + ½ (-9.8) t²

4.9 t² - 73.9 t - 140 = 0

Solve with quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8

t = -1.7, 16.8

Since t can't be negative, t = 16.8.  Rounding to 2 sig-figs, the projectile lands after 17 seconds.

Answer:

4 m/s2 in negative acceleration

Explanation:

If An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds, it is 4 m/s2 in negative acceleration.

Hope this helps!

Answer:

[tex]-4\frac{m}{s^2}[/tex]

Explanation:

The object changes its speed over some time, this means that there is an acceleration.

It has a uniformly accelerated movement.

The Formula for finding speed in a uniformly accelerated motion is

[tex]a=\frac{V_{f}-V_{o}}{t}[/tex]

[tex]V_{o}= 65\frac{m}{s}\\V_{f}= 25\frac{m}{s} \\t= 10s[/tex]

Replace

[tex]a=\frac{(25-65)\frac{m}{s} }{10 s}\\ a=\frac{-40\frac{m}{s} }{10s}\\a= -4\frac{m}{s^2}[/tex]

Acceleration gives us a negative value this means that it is slowing.

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude

[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]

We have

[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]

where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so

[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so

[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]

Then the magnitude of the particle's acceleration is

[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]

The magnitude of the acceleration of the particle is approximately 70.55 m/s^2, calculated by using the formulas for combined radial and tangential acceleration in circular motion.

In this physics problem, the particle not only moves around in a circle but is also experiencing an increase in speed which is a case of combined radial and tangential acceleration. Radial acceleration, known as centripetal acceleration (ar), is the result of the change in direction of the velocity vector, while tangential acceleration (at) comes from changes in speed.

The total acceleration of an object in circular motion is given by:

a = sqrt((ar^2) + (at^2))

Centripetal acceleration can be calculated using the formula ar = v^2 / r, where: v = speed (37.2 m/s), r = radius of the circle (21 m). This gives us ar = (37.2^2) / 21, which approximately equals 66.62 m/s^2.

The tangential acceleration is given in the problem: at = 23.1 m/s^2.

We therefore calculate the total acceleration using the formula above which gives us:

a = sqrt((66.62^2) + (23.1^2)) which approximately equals 70.55 m/s^2.

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Answer:

Option (A)

Explanation:

A ball bearing is a device which is use to reduce the friction.

The outer rim of the bearing is fixed with the part of machine and inner rim is fitted into shafts. Now the shafts rotates and only the small spheres in the bearing will rotate. The friction can be further reduced by apply the oil or grease to the bearing.

Answer:

10.19 m

Explanation:

Energy is conserved, so elastic energy stored in spring = gravitational energy of block.

1/2 kx² = mgh

h = kx² / (2mg)

h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)

h = 10.19 m

Answer:

0.0018 V

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the particle is equal to the electric potential energy lost:

[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=3.8\cdot 10^{-6} kg[/tex] is the mass of the particle

[tex]v=36 m/s[/tex] is the final speed of the particle

q = -2.7 C is the charge

[tex]\Delta V[/tex] is the potential difference between the two points

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V[/tex]

The particle has been accelerated by this potential difference: since it is a negative charge, it means that the particle has moved from a point at lower potential towards a point of higher potential.

So, since the initial point is A and the final point is B, the result is

[tex]V_B - V_A = 0.0018 V[/tex]

Answer:

a)

16.33 Ω

b)

45009.18 A

Explanation:

a)

L = length of the line = 935 km = 935000 m

d = diameter of the line = 3.50 cm = 0.035 m

ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m

Area of cross-section of the line is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.035)²

A = 0.000961625 m²

Resistance of the line is given as

[tex]R=\frac{\rho L}{A}[/tex]

inserting the values

R = (1.68 x 10⁻⁸) (935000)/(0.000961625)

R = 16.33 Ω

b)

V = potential difference across the line = 735 kv = 735000 Volts

i = current carried by the wire

Using ohm's law, current carried by the wire is given as

[tex]i=\frac{V}{R}[/tex]

i = 735000/16.33

i = 45009.18 A

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

Answer:1074.26

Explanation:just got it right on my accelus

Answer:

The cannon recoils with a force of 332.5 N

Explanation:

By Newton's third law Recoil force on cannon = Force in shell.

Force in shell = Mass of shell x Acceleration of shell

Mass of shell = 3.5 kg

Acceleration of shell = 95 m/s²

Force in shell = 3.5 x 95 = 332.5 N

Recoil force on cannon = 332.5 N

So, the cannon recoils with a force of 332.5 N

Answer with Explanation:

since the two blocks move (stick) together, the collision is inelastic, which does not conserve kinetic energy.  So do not use kinetic energy consideration.

Fortunately, in such a situation, momentum is still conserved.

Momentum of 1.0 kg block

= 1.0 * 3.0 = 3.0 kg-m/s

Momentum of second block

= 1.0 * 1.0 = 1.0 kg-m/s

Total mass after collision = 1.0+1.0 = 2.0 kg

Common velocity after collision

= total momentum / total mass

= (3.0+1.0)/2.0 = 2.0 m/s

Answer:A

Explanation:

Initial velocity, u = 2m/s

Final velocity, v = 4m/s

Distance covered, s = 12m

Acceleration, a = ?

Using

v² = u² + 2as

2as = v² - u²

a = v²-u²/2s

a = 4²-2²/2 x 12

a = 16-4/24

a = 12/24

a = 0.5m/s²

Answer:

0.96 m/s

Explanation:

Consider the motion of the ball before collision with the block

h = height from which the ball is dropped = length of the wire = 1.01 m

m = mass of the ball = 1.67 kg

v = speed of the ball just before collision with the block = ?

Using conservation of energy for the ball

Kinetic energy of the ball at the bottom = Potential energy of the ball at the top

(0.5) m v² = mgh

(0.5) v² = gh

(0.5) v² = (9.8) (1.01)

v = 4.45 m/s

consider the collision between the ball and the block :

m = mass of the ball = 1.67 kg

v = velocity of the ball just before collision with the block = 4.45 m/s

v' = velocity of the ball just after collision with the block

M = mass of the block = 2.59 kg

V = velocity of the block just before collision with the ball = 0 m/s

V' = velocity of the block just after collision with the ball

Using conservation of momentum

mv + MV = mv' + MV'

(1.67) (4.45) + (2.59) (0) = 1.67 v' + (2.59) V'

7.43 = 1.67 v' + (2.59) V'

[tex]V' = \frac{(7.43 - 1.67 v')}{2.59}[/tex]                                eq-1

Using conservation of kinetic energy

(0.5) mv² + (0.5) MV² = (0.5) mv'² + (0.5) MV'²

mv² + MV² = mv'² + MV'²

(1.67) (4.45)² + (2.59) (0)² = 1.67 v'² + (2.59) V'²

using eq-1

(1.67) (4.45)²  = 1.67 v'² + (2.59) ((7.43 - 1.67 v')/2.59)²

v' = - 0.96 m/s

the negative sign indicates the direction which is opposite to its direction before colliding with the block.

To find the velocity of the ball just after the elastic collision, apply the formula for one-dimensional elastic collision. Given that the ball initially fell from a certain height, calculate its initial velocity using the conservation of energy principle. Insert these results, along with the given masses, into the elastic collision formula to achieve the answer.

This question explores the principles of conservation of momentum and the properties of an elastic collision. Given that this collision is perfectly elastic, we can use the formula for an elastic collision in one-dimension:

v1' = ((m1 - m2) / (m1 + m2)) * v1 + ((2 * m2) / (m1 + m2)) * v2

where v1' is the velocity of the ball after collision, m1 and m2 are the masses of the ball and the block respectively, v1 and v2 are the velocities of the ball and block before the collision. The ball was initially dropped from a height, so we'd first need to calculate its velocity just before collision using the conservation of energy principle:

v1 = sqrt(2* g * h)

where g is the acceleration due to gravity and h is the height (length of the wire). After figuring out the value of v1, we just need to plug the values into the first equation to find v1' - velocity of the ball after collision.

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Answer:

2.4 x 10^21

Explanation:

V = 9 V, R = 1.4 ohm, t = 1 minute = 60 second

Use Ohm's law

V = I R

I = V / R

I = 9 / 1.4

I = 6.43 A

Now use Q = I t

Q = 6.43 x 60 = 385.7 C

Number of electrons passing in 1 minute , n

= total charge in one minute / charge of one electron

n = 385.7 / (1.6 x 10^-19) = 2.4 x 10^21

Answer:

(e) 98,1 KJ

Explanation:

The engine produces 19%; it means, it rejects 81% of energy. ⇒ 81/19=4.26 times.

The engine produces 23 kJ; it means it rejects 23 * 4.26 = 98.05263 kJ

Answer:

320 mm

Explanation:

The equivalent length for a minor loss is:

Le = K D / f

where K is the loss coefficient, D is the diameter, and f is the friction factor.

Le = (0.4) (20 mm) / 0.025

Le = 320 mm

The equivalent length (Le) for the elbow with a loss coefficient (K) of 0.4, a pipe with a friction factor of 0.025, and a diameter of 20 mm, is 320 mm.

To calculate the equivalent length (Le) for the elbow in terms of pipe diameter, we use the relationship between the loss coefficient (K), the friction factor (f), and the diameter (D) of the pipe, given by the equation Le = (K * D) / f. In this case, the loss coefficient for the elbow is 0.4, the friction factor is 0.025, and the diameter of the pipe is 20 mm. Plugging these values into the equation, we get:

Le = (0.4 * 20 mm) / 0.025 = 320 mm.

Therefore, the equivalent length for the elbow is 320 mm.

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

[tex]L_2 - L_1 = 10 Log \frac{I_2}{I_1}[/tex]

so here we need to find the ratio of two intensity

it is given as

[tex]Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = \frac{60 - 30}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = 3[/tex]

now we have

[tex]\frac{I_2}{I_1} = 10^3[/tex]

so it is

d) 1000 times

Answer:

0.8 m

Explanation:

For "6 N" force :

F = magnitude of the force = 6 N

r = moment arm = 0.4 m

Torque due to "6 N" force is given as

τ = r F

τ = (0.4) (6)

τ = 2.4 Nm

For " 15 N" force :

F' = magnitude of the force = 15 N

r' = moment arm = ?

τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm

Torque due to "15 N" force is given as

τ' = r' F'

12 = r' (15)

r' = 0.8 m

So the moment arm for "15 N" force is 0.8 m

Answer:

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

[tex]tan\theta_i = \frac{0.7}{0.98}[/tex]

[tex]\theta_i = 35.5^0[/tex]

now by Snell'a law at that interface we have

[tex]\mu_1 sin\theta_i = \mu_2 sin \theta_r[/tex]

now we will have

[tex]1.52  sin35.5 = 1.40 sin\theta_r[/tex]

[tex]\theta_r = 39.12^0[/tex]

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

[tex]\mu_2 sin\theta_i' = \mu_{air} sin\theta[/tex]

[tex]1.40 sin39.12 = 1 sin\theta[/tex]

[tex]\theta = 62.1^0[/tex]

The problem involves applying Snell's law twice, once for the brine-to-oil interface and then for the oil-to-air interface, with angles calculated from the normal lines. Severally, the use of the Pythagorean theorem gives the angle in brine, Snell's law then gives the angle in oil, and finally in the air. The desired angle is found as an augmentation to 90 degrees since it should be calculated from the axis, not the normal line.

Firstly we need to calculate the angle that the ray makes with the norm or perpendicular line to the brine-oil interface. This can be done using the Pythagorean theorem, given that we have the height of the brine (0.98m), and the horizontal distance from the axis to the point of intersection (0.7m). The angle is then the arcsin of the opposite side (0.7m) over the hypotenuse (√[(0.7m)2+(0.98m)2]) in radians. (Remember that inverse sin is used to calculate angles). Let's call this angle θ1.

Using Snell's law (n1sinθ1 = n2sinθ2), where the indices of refraction n1 and n2 are for brine (1.52) and oil (1.40) respectively, we can figure out θ2, the angle that the ray makes with the normal in the oil part. The key detail is that angles are always measured from the normal line, which is perpendicular to the interfaces.

Finally, similar calculations will grant what is to be found - the angle the ray makes with the vertical axis in the air (with the index of refraction n=1). This is θ3 calculated from the equation n2sinθ2 = n3sinθ3, where θ2 is the angle that we just found and n2 is the index of refraction for oil. You need to remember that Snell's law applies for any two mediums with a clear border in between through which the light passes.

Finally, the desired angle between the ray's line in the air and the vertical axis will be 90° - θ3 (since θ3 was an angle with the normal line, which is defined as perpendicular to the axis).

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Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:  

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

[tex]T = \frac{2 u Sin\theta }{g}[/tex]

[tex]T = \frac{2 \times 20 \times Sin35 }{9.8}[/tex]

T = 2.34 second

(b) The formula for the maximum height is given by

[tex]H = \frac{u^{2} \times Sin^{2}\theta }{2g}[/tex]

[tex]H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}[/tex]

H  = 6.71 m

(c) The formula for the range is given by

[tex]R = \frac{u^{2} \times Sin 2\theta }{g}[/tex]

[tex]R = \frac{20^{2} \times Sin 2 \times 35}{9.8}[/tex]

R = 38.35 m

(d) It hits with the same speed at the initial speed.

Answer:

Force required is 4387 N in the opposite direction of motion.

Explanation:

We have equation of motion v² = u² + 2as

      v = 0m/s , u = 15 m/s, s = 31.8 m

  Substituting

         0² = 15² + 2 x a x 31.8

          a = -3.54 m/s²

So, deceleration = 3.54 m/s²

Force = Mass x Acceleration

          = 1240 x -3.54 = -4387 N

So force required is 4387 N in the opposite direction of motion.

Answer:

Wavelength: the distance between two crests of an electromagnetic wave

Frequency: the rate at which electromagnetic waves oscillate

Explanation:

For a transverse wave (such as the electromagnetic waves), we can define the following quantities:

Wavelength: it corresponds to the distance between two consecutive crests (or between two consecutive troughs) of the wave

Frequency: the number of complete oscillations of the wave per unit time

Period: it is the time it takes for the wave to complete one oscillation - it is equal to the reciprocal of the frequency

Amplitude: it is the maximum displacement of the wave, measured with respect to its equilibrium position

Answer:

2.0 s, 200 m

Explanation:

Time to hit the ground depends only on height.  Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.

Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance).  So it will travel a horizontal distance of 200 m.

Answer:

1) 2 seconds

2) 200 m

Explanation:

1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]

  Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]

  Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]

s = speed

t = time

h = height

Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,

Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]

i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s

Time taken to land = 2.0 s

2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m

   Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]

Therefore, since distance traveled is directly proportional to speed,

Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]

i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]

    [tex]2S_{1}[/tex] = 2 x 100 m = 200 m

Distance traveled = 200 m