Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

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Answer:

Tension in the vertical rope: approximately [tex]613 \; {\rm N}[/tex].

Tension in the horizontal rope: approximately [tex]3.74 \times 10^{3}\; {\rm N}[/tex].

Assumption: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

Explanation:

Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:

[tex]\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is in the clockwise direction.

The weight of the beam ([tex]m = 116\; {\rm kg}[/tex]) would be:

[tex]\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of [tex](7.65\; {\rm m}) / 2[/tex] from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is also in the clockwise direction.

The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:

[tex]\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

Note the angle between the direction of this tension and the beam is [tex](90^{\circ} - \theta) = 32.3^{\circ}[/tex]. This force is applied  [tex](7.65\; {\rm m}) / 2[/tex] from the pivot. Hence, achieving that torque of [tex]7.645 \times 10^{3}\; {\rm N \cdot m}[/tex] would require:

[tex]\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}[/tex].

The elastic potential energy of the block-spring system is 0.906 J.

Velocity of the block , v = 0.83 m/s.

Elastic potential energy is the energy stored in a stretched or compressed elastic material.

The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J

The elastic potential energy of the spring is converted to kinetic energy of the block.

1/2 mv² = 0.906 J

where v is velocity

v = √(0.906 * 2)/2.6

v = 0.83 m/s.

In conclusion, elastic potential energy is present in compressed or stretched elastic materials.

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Answer:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

Accuracy means the state of being accurate, without any mistakes and the results should be 100% true.

Whereas, precision means, approximately true or almost true.

Hope it help you

Answer:

Maximum height = 2040 m

Explanation:

We can solve the problem using kinematics.

Consider the vertical motion of the object and use the equation:

[tex]\boxed{v^2 = u^2 + 2as}[/tex]

where:

• v = final velocity      (0 m/s, because when the object is at max. height, it has no vertical velocity)

• u = initial velocity    (400sin30° m/s ⇒ vertical component of 400 m/s at 30° to horizontal)

• a = acceleration      (-9.81 m/s²; considering upward acceleration to be negative)

• s = displacement    (? m; this represents the max. height of the object),

Substitute the values into the equation and solve for s :

[tex]0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)[/tex]

⇒ [tex]2(9.81)(s) = (400 sin (30 \textdegree))^2[/tex]

⇒  [tex]s = 2040 \space\ m[/tex]     (3 s.f.)

The required horsepower engine is 32 horsepower.

The force of the 30 skiers is calculated as follows:

Mass of the skiers = 30 * 67kg = 2010 kg

Net force acting on the skiers along the x-axis

where f is the frictional force

The kinetic frictional force, f = μN

where

μ = The coefficient of the kinetic friction

N = normal reaction

Net force acting on the skiers along y axis, the

Fy = ma

N = mg cos θ

Substituting for N above

f = μk mg cos θ

Substituting for f in  (1)

F = mg sin θ + μk mg cos θ

F = mg(sinθ + μk cos θ)

Work done by the engine in pilling up the skiers, W = Fx

W = mg ( sinθ + μk cos θ)x

x = 300 m

W = (2010 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (300 m)

Work done, W = 2.86 * 10⁶ J

Time taken, t = 2.0 * 60sec = 120 s

1 horsepower = 746 W

Power = 2.86 * 10⁶/ 120 * 1/746

Power = 31.9 horsepower

In conclusion, the power of the engine is the ratio of the work done and time taken.

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Hi there!

Part A.

To solve this part, all we need to do is a summation of vertical forces.

We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)

These sum to zero because the beam is not accelerating vertically.

[tex]\Sigma F = -F_g - T + P_y = 0[/tex]

[tex]P_y = F_g + T[/tex]

The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.

[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]

Part B.

We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)

- The tension of the horizontal cable acting at the end of the rod (CCW, at L)

- The force of tension in the rope holding the box (CC, at 3L/4)

Since the rod is not rotating, these torques sum to zero.

The equation for torque is:
[tex]\tau = r \times F[/tex]

This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.

Now, let's find the torque from all three of these forces.

- Force of gravity at center:

The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.

[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]

- Tension of horizontal cable:

The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.

[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]

- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]

Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]

Part C.

This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.

We only have two:
- The horizontal tension in the cable to the left (1847.38 N)

- The horizontal force exerted by the pivot on the beam to the right

These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]

**to the right

Answer:

a) 960.4 N

b) T= 5/4 Mg CotanΘ

c) 1847. 38

Explanation:

a) Py= 2Mg

=2(49 x 9.8)

= 960.4

b) T= (Mg x 1/2 x cos Θ + Mg x 3/4 x cos Θ) / sin Θ

T= 5/4 X Mg cotanΘ

c) T= (5/4) x (49 x 9.8) cotan (18)

T= 1847.37954

= 1847.38

A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) Collecting dew from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

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Big bang happened about 13.7 billion years ago in our universe.

According to the big bang theory, about 13.7 billion years ago, an explosive expansion began, expanding our universe outwards faster than the speed of light.

According to the flat model, the universe is infinite and will continue to expand forever because the universe is expanding.

Cosmic background radiation is a weak radio-frequency radiation that is traveling through outer space in every direction. It is the residual radiation of the big bang, when the universe was very hot.

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe has all over radiation which is called the “cosmic microwave background".

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Answer:

pictures please

Explanation:

I need a picture so I can tell you

The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                                    [tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]

                                 [tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]

                                 [tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.

We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.

                              [tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]

                                [tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]

                                     [tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]

Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.

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The spring constant, k, in newtons per meter is 1,955.9 N/m.

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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(a) The maximum height reached by Jane is 1.8 m.

(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

h = (6²)/(2 x 9.8)

h = 1.84 m

Assuming Jane to be in simple harmonic motion, the time of motion is calculated as;

T =  2π√(L/g)

where;

Thus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

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Answer:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

All of the possible metalloids are:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)

In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

The term acceleration has to do with the rate at which velocity changes with time.

We have to take the moments at the tipping point of rotation as follows;

Clockwise moment = Anticlockwise moment

Hence;

F₂ * 1.58 m = F₁ * 0.67 m

The weight at half the width= 30 cm or 0.3 m

Height of refrigerator = 158 cm 0r 1.58 cm

Then;

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

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B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

Fₙ = mg + mv²/r

where;

Fₙ = (2,000 x 9.8) + (2,000 x 18²)/24

Fₙ = 46,600 N

Fₙ =  4.7 x 10⁴ N

Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

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The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.

we know:-

Mass = 69 kg

Speed = 3.2 m/s

Coefficient of friction ( ratio of friction force and normal force ) = 0.70

Acceleration due to gravity, g = 9.8 m/s^2

(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:

[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]

      [tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]

      [tex]= 353.28[/tex] Joules

Mechanical energy = 353.28 Joules

(b) To determine how far (distance) the runner slide:

acceleration = ug

                     [tex]= 3.2[/tex] × [tex]9.8[/tex]

                     [tex]= 31.36[/tex] [tex]m/s^{2}[/tex]

distance ,

[tex]V^{2} = U^{2} + 2aS[/tex]

[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]

[tex]10.24 = 62.72 S[/tex]

[tex]S = {\frac{10.24}{62.72} }[/tex]

Distance, S = 0.16 m  

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(a) The normal force exerted by the road on the car is 8,506.2 N.

(b) The normal force exerted by the car on the driver is 394.9 N.

(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.

The normal force exerted by the road on the car is calculated as follows;

Normal force = weight of the car - centripetal force of car

Weight of the car = (1400 x 9.8) = 13,720 N

Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N

Normal force = 13,720 N -  5,213.8 N

Normal force = 8,506.2 N

Normal force = weight of driver - centripetal force of driver

Weight of driver = (65 x 9.8) = 637 N

Centripetal force of driver = (65² x 18²)/(87) = 242.07 N

Normal force =  637 N - 242.07 N = 394.9 N

N = mg - mv²/r

0 = mg - mv²/r

mv²/r = mg

v²/r = g

v² = rg

v = √rg

v = √(87 x 9.8)

v = 29.2 m/s

Thus, the normal force exerted by the road on the car is 8,506.2 N.

The normal force exerted by the car on the driver is 394.9 N.

The speed of the car at which the normal force on driver is zero is 29.2 m/s.

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Hello!

a)

To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.

Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)

i = Current in wire ('I' A)
dl = differential length element

r = distance from wire to point P ('R' m, remains constant!)

We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

First, let's mind the cross-product.

The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.

Therefore, we can disregard the cross-product since sin(90) = 1.

Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:

[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]

We have a semi-circle, so we are integrating from 0 to π radians.

[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]

Simplifying to make the integral easier, we can take constants out of the integral.

[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]

Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]

b)

Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.

The volume of the helium balloon in order to lift the weight is 17,760m³.

To find the answer, we need to know about the buoyant force.

     => V= 3179/0.179 = 17,760m³

Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.

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Answer:

Higher levels of organization are built from lower levels.

Molecules combine to form cells.

Cells combine to form tissues.

Tissues combine to form organs.

Organs combine to form organ systems,

and organ systems combine to form organisms.

Explanation:

Hope it helps.

Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

[tex]\Sigma \tau = 0[/tex]

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]

Doing the summation using their respective lever arms:

[tex]0 = L Tsin\theta - dF_g[/tex]

[tex]dF_g = LTsin\theta[/tex]

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]

Now, let's solve for 'T'.

[tex]T = \frac{dMg}{Lsin\theta}[/tex]

Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]

Answer:  Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure. The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). Then, at an angle 135.48°, the planet Y rotated through during this time.

Explanation: To find the answer, we need to know about the Kepler's third law of planetary motion.

                                      T² ∝ r³

                 [tex]\frac{r_1}{r_2}=\frac{4}{5} \\\frac{T_1}{T_2} =(\frac{r_1}{r_2})^{3/2}\\[/tex]

                  [tex]T=\frac{2\pi }{w}[/tex] where, w is the angular velocity.

                           [tex]\alpha[/tex] =wt.

                   [tex]\frac{T_x}{T_y}=\frac{w_y}{w_x}\\thus,\\\frac{w_y}{w_x}=(\frac{r_1}{r_2})^{3/2}[/tex]

                     [tex]\frac{\alpha _y}{\alpha _x} = (\frac{r_1}{r_2})^{3/2}\\where,\\\alpha _x=\frac{88^0}{5 yrs} because,\\here, angle \beta_x =88^0.\\[/tex]

                     [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} )^{3/2}=\frac{135.48^0}{5yrs} .\\thus,\\\beta _y=135.48^0.[/tex]

Thus, we can conclude that the angle rotated by planet Y during 5 yrs will be 135.48 degrees.

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The planet Y then rotated through at this time at an angle of 135.48°.

In order to understand the solution, we must be familiar with Kepler's third law of planetary motion.

                              T² ∝ r³

                                  [tex]\frac{R_1}{R_2}=\frac{4}{5} \\\frac{T_1}{T_2}=(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

                                [tex]T=\frac{2\pi }{w}[/tex]

where w is the angle of rotation per time.

                                     [tex]\alpha[/tex]=wt.

                                   [tex]\frac{w_y}{w_x} =(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

                                   [tex]\frac{\alpha _y}{\alpha _x}=(\frac{4}{3} ) ^{\frac{3}{2} } , where\\\alpha _x=\frac{88}{5yrs} \\[/tex]

                           [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} ) ^{\frac{3}{2} }=\frac{135.48}{5yrs}[/tex]

Thus, we may infer that planet Y will revolve at an angle of 135.48 degrees during the course of five years.

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(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d -  I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L =  (μI/2π) x (I/2d + I/d)

F₃/L =  (μI/2π) x (3I/2d)

F₃/L =  3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

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The the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

The lead weight will fall under the influence of gravity with magnitude of 9.8 m/s².

t = √2h/g

where;

t = √((2 x 6.004)/9.8)

t = 1.11 s

Thus, the the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

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(a) The location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

(b) The magnitude and direction of the electric field at the top corner due to the two charges at the base is  1.732 kq/a².

The electric field is zero at the center of the equilateral triangle whose magnitude is equal to √3a/6.

E = E₁ + E₂

where;

E = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]

E = kq/a²[2(sin 60j)] = 1.732 kq/a²

Thus, the location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

The magnitude and direction of the electric field at the top corner due to the two charges at the base is  1.732 kq/a².

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From the calculation, the normal force is 6161.2 N.

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

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Answer:

D

Explanation:

momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.

and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.

so, the sign is only for the direction.

the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.

It is possible that momentum can be negative if an object is moving backward. The correct option is D.

A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.

When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.

Thus the correct option is D.

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