How does universal gravitation affect you and objects around you?

Answer:

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

[tex]tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\[/tex]

Then we find the angle β of the displacement vector

[tex]tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\[/tex]

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

[tex]W=F*d*cos (theta)[/tex]

The absolute value of F will be:

[tex]F=\sqrt{8^{2}+1^{2}  } \\F= 8.06 N[/tex]

The absolute value of d will be:

[tex]d=\sqrt{(6 )^{2}+(2)^{2}  } \\d= 6.32m\\[/tex]

Now we have:

[tex]W=8.06*6.32*cos(25.56)\\W=46 J[/tex]

b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

a) the force from the engine is greater than all the forces of friction.  --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,

a = 0

and so Newton's second law becomes

[tex]\sum F = ma = 0[/tex]

where [tex]\sum F[/tex] is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero.  --> TRUE, for what we said at point A)

c) the inertia is changing.  --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car.  --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

[tex]F_f = \mu mg[/tex]

where [tex]\mu[/tex] is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

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Answer:

The work done by the force of gravity, W = 0 J

Explanation:

Given data,

The weight of the box, w = 1500 N

Therefore the gravitational force acting on the box is , F = 1500 N

The displacement of the box S = 0

The work done is defined as the product of force acting on the body to the displacement it caused.

The formula for work done is,

                                           W = F x S

                                                = 1500 N x 0

                                                = 0

Hence. the work done by the force of gravity is, W = 0 J

The work done by the force of gravity on a box weighing 1500 N depends upon the distance the box is moved. If we assume a free fall situation, the work done would be calculated using the formula: W = 1500N * h, where h is the height from which the box falls.

The calculation of the work that the force of gravity does on an object comes down to two important factors: the force of gravity acting on the object, represented as weight, and the distance the object is moved. In this case, the box weighs 1500 N, which is the force of gravity acting on it. If we assume that the box is in free fall, i.e., it's falling to the ground under gravity, we can calculate the work done by the force of gravity.

To calculate this, we need to know the height from which the box is falling (let's denote that as h). The work done, according to the work-energy principle, is equal to the force times the distance covered and the cosine of the angle between force and distance vectors. Considering the angle between the gravity force and the displacement is 0 degrees in a free fall, the work done by gravity is W = 1500N * h * cos(0) = 1500N * h. Here, cos(0) is equal to 1.

As you can see, the work done by gravity relies heavily on the distance/height textbox falls. If the box doesn't move, then no work is done by gravity even though the force of gravity exists, because work is dependent on displacement.

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The distance covered is 30.2 m

Explanation:

The motion of Cynthia is a uniformly accelerated motion (constant acceleration), so we can find the distance covered by using the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem,

u = 0 (since Cynthia started from rest)

t = 3.2 s

[tex]a=5.9 m/s^2[/tex] is her acceleration

Substituting,

[tex]s=0+\frac{1}{2}(5.9)(3.2)^2=30.2 m[/tex]

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Using the equation of motion S = ut + ½at2, we calculated that Susan traveled 30.216 meters during the initial stage of her downhill luge, where she accelerated at 5.9 m/s2 for 3.2 seconds.

To calculate how far Susan traveled during the initial stage of her downhill luge, where she accelerated from rest at 5.9 m/s2 for 3.2 seconds, we use the equation of motion for constant acceleration:

S = ut + ½at2

Here, S represents the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Since Susan starts from rest, her initial velocity u = 0. We plug in the acceleration a = 5.9 m/s2 and time t = 3.2 s into the equation:

S = (0)(3.2) + ½(5.9)(3.2)2

S = 0 + ½(5.9)(10.24)

S = 30.216 m

Therefore, Susan traveled 30.216 meters during the initial stage of her downhill luge.

Answer:

The change in momentum of the car is  30,000 kg m/s                  

Explanation:

Given,

The force exerted on the car to slow down, F = 7500 N

The time period of force, t = 4 s

The rate of change of momentum of the object is equal to the force acting on it.

Therefore,

                                    (mv - mu) / t = F

Where v and u are the final and initial velocity of the car. The change in momentum of the car,

                                    mv - mu = F x t

                                                   = 7500 x 4

                                                   = 30,000 kg m/s

Hence the change in momentum of the car is  30,000 kg m/s                  

In a collision between two 300 kg spaceships, one at rest and the other moving at 6 m/s, the final velocity after they stick together is 2 m/s.

When two objects with masses of 300 kg collide and stick together, their momentum is conserved. Utilizing the principle of conservation of momentum, we can calculate their final velocity after the collision.

In this case, since both spaceships have equal masses and one is at rest while the other moves with a known speed, the final velocity after the collision would be 2 m/s.

The maximum tension in the rope during the swing is the sum of the gravitational force and the centripetal force, which can be calculated with the given mass, rope length, and speed, resulting in a tension of 1602.48 N.

To calculate the maximum tension in the rope during the swing, we need to consider the forces acting on the man at the lowest point of the swing. These forces are the gravitational force (weight) and the centripetal force required to maintain the circular motion. The tension in the rope is the sum of these two forces.

The gravitational force (Fg) on the man can be calculated using the equation Fg = m  imes g, where m is the man's mass (88.00 kg) and g is the acceleration due to gravity (9.81 m/s2). The centripetal force (Fc) required for circular motion can be calculated using the equation Fc = m imes v2 / r, where v is the man's speed (10.20 m/s) at the bottom of the swing and r is the radius of the swing, which is equal to the length of the rope (12.0 meters).

Therefore, the maximum tension (T) in the rope would be:

T = Fg + Fc

T = (88.00 kg  imes 9.81 m/s2) + (88.00 kg  imes (10.20 m/s)2 / 12.0 m)

Calculating each part:

Fg = 863.28 N (2 decimal places)

Fc = 739.20 N (2 decimal places)

Thus, the maximum tension in the rope would be:

T = 863.28 N + 739.20 N

T = 1602.48 N (rounded to two decimal places)

Tension, centripetal force, and gravitational force are crucial components in determining the tension in the rope during circular motion.

The velocity of the wave is 40 m/s

Explanation:

The relationship between the velocity of a wave and its frequency and wavelength is given by

[tex]v=f \lambda[/tex]

where

v is the velocity

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the ocean wave in this problem, we have:

[tex]\lambda = 10 m[/tex] (wavelength)

f = 4.0 Hz (frequency)

Therefore, its velocity is

[tex]v=(4.0)(10)=40 m/s[/tex]

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The velocity of an ocean wave with a wavelength of 10 m and a frequency of 4.0 Hz is 40 m/s.

The subject of this question is Physics, and more specifically, it deals with the topic of wave motion and wave velocity. The velocity of a wave can be calculated using the equation v = f × lambda, where v is the velocity of the wave, f is its frequency, and lambda is its wavelength.

Given that the wavelength lambda is 10 m and the frequency f is 4.0 Hz, we can substitute these values into the equation to find the wave velocity:

v = f × lambda = 4.0 Hz × 10 m = 40 m/s.

Therefore, the velocity of the ocean wave is 40 meters per second (m/s).

The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.

In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.

In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.

Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.

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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.

Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:

|a| = |v|²/r

From the problem, we have two cases:

[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]

Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]

For case A:

[tex]|a_A| = |v_A|^2 / |r_A|[/tex]

For case B:

[tex]|a_B| = |v_B|^2 / |r_B|[/tex]

Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:

[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]

Simplifying this, we get:

[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]

Therefore, the acceleration in case A is twice that of case B.

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

[tex]v_x = u cos \theta[/tex]

where

u = 8.3 m/s is the initial velocity of the balloon

[tex]\theta=27^{\circ}[/tex] is the angle of projection

Substituting,

[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]

b)

The vertical component of the initial velocity of a projectile is given by

[tex]u_y = u sin \theta[/tex]

where

u is the initial velocity

[tex]\theta[/tex] is the angle of projection

Here we have

u = 8.3 m/s

[tex]\theta=27^{\circ}[/tex]

Substituting,

[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]

c)

The vertical component of the velocity of the balloon follows the suvat equation

[tex]v_y = u_y - gt[/tex]

where

[tex]v_y[/tex] is the vertical velocity at time t

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

[tex]v_y = 0[/tex]

So we get:

[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]

d)

The maximum height of the balloon can be calculated using the suvat equation:

[tex]s=u_y t - \frac{1}{2}gt^2[/tex]

where

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

[tex]t=\frac{2u_y}{g}[/tex]

where

[tex]u_y[/tex] is the initial vertical velocity

[tex]g[/tex] is the acceleration of gravity

Here we have

[tex]u_y = 3.8 m/s[/tex]

[tex]g=9.8 m/s^2[/tex]

Substituting,

[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

[tex]d=v_x t[/tex]

where

[tex]v_x[/tex] is the horizontal velocity

t is the time of flight

Here we have

[tex]v_x = 7.4 m/s[/tex]

[tex]t = 0.78 s[/tex]

Substituting,

[tex]d=(7.4)(0.78)=5.77 m[/tex]

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Answer:

The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

[tex]\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}[/tex]

[tex]\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}[/tex]

[tex]f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)[/tex]

[tex]\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)[/tex]

[tex]\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

[tex]v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

Substitute the given values in the formula,

[tex]v_{s}=343+\frac{262}{271}(343-0)[/tex]

[tex]v_{s}=343+0.966(343)[/tex]

[tex]v_{s}=343-331.33[/tex]

[tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Therefore, The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

The correct answer is a. potential energy. At the top of a hill, right before it plunges downward, a roller coaster car has potential energy due to its position relative to the ground. This potential energy is converted into kinetic energy as the car moves downward and gains speed.

Answer:

Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.

The kinetic theory of matter states that all matter is made up of tiny particles that are constantly moving. The faster the particles are moving, the higher the temperature of the matter.

When a liquid evaporates, the fastest-moving particles are the ones that are most likely to escape from the surface of the liquid. This leaves behind the slower-moving particles, which means that the average temperature of the liquid decreases.

Another way to think about it is that evaporation is a cooling process because it removes the fastest-moving particles from the liquid. These particles have the most kinetic energy, so their removal lowers the average kinetic energy of the remaining particles and thus lowers the temperature of the liquid.

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Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

Answer:

180

Explanation:

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second. The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius.

Answer:

160kg

Explanation:

Given m= 95kg g= 1.6345m/s²

Unknown Fg= ?

Formula Fg=mg

Fg=(95)(1.6345)

Fg= 155.2775

Fg=160kg

a) The frequency of the photon is [tex]7.16\cdot 10^{20}Hz[/tex]

b) The wavelength of the photon is [tex]4.19\cdot 10^{-13} m[/tex]

c) The wavelength of the photon is about 100 times larger than the nuclear radius

Explanation:

a)

The energy of a photon is given by

[tex]E=hf[/tex] (1)

where:

[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant

f is the frequency of the photon

The photon in this problem has an energy of

[tex]E=2.5 MeV = 2.5\cdot 10^6 eV[/tex]

And keeping in mind that

[tex]1eV = 1.6\cdot 10^{-19} J[/tex]

we can convert to Joules:

[tex]E=(2.5\cdot 10^6)(1.9\cdot 10^{-19})=4.75\cdot 10^{-13} J[/tex]

And now we can use eq.(1) to find the frequency of the photon:

[tex]f=\frac{E}{h}=\frac{4.75\cdot 10^{-13}}{6.63\cdot 10^{-34}}=7.16\cdot 10^{20}Hz[/tex]

b)

The wavelength of a photon is related to its frequency by the equation

[tex]c=f\lambda[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the photon in this problem,

[tex]f=7.16\cdot 10^{20}Hz[/tex]

Re-arranging the equation, we find its wavelength:

[tex]\lambda=\frac{c}{f}=\frac{3\cdot 10^8}{7.16\cdot 10^{20}}=4.19\cdot 10^{-13} m[/tex]

c)

The size of the nuclear radius is approximately

[tex]d \sim 10^{-15} m[/tex]

While we see that the wavelength of this photon is

[tex]\lambda=4.19\cdot 10^{-13} m[/tex]

Therefore, the ratio between the wavelength of the photon and the nuclear radius is

[tex]\frac{\lambda}{d}=\frac{\sim 10^{-13}}{\sim \cdot 10^{-15}}=100[/tex]

So, the wavelength of the photon is approximately a factor 100 times larger than the nuclear radius.

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1) Mass of the continent: [tex]2.2\cdot 10^{21} kg[/tex]

2) Kinetic energy: 1118 J

3) Speed of the jogger: 5.3 m/s

Explanation:

1)

First of all, we calculate the volume of the continent. It is a slab of side

[tex]L=5100 km = 5.1\cdot 10^6 m[/tex]

and thickness

[tex]t=30 km = 3.0\cdot 10^4 m[/tex]

So its volume is

[tex]V=tL^2=(3.0\cdot 10^4)(5.1\cdot 10^6)^2=7.8\cdot 10^{17} m^3[/tex]

The density of the slab is

[tex]\rho = 2850 kg/m^3[/tex]

Therefore, we can calculate the mass using the relationship

[tex]\rho = \frac{m}{V}[/tex]

where m is the mass. And solving for m,

[tex]m=\rho V=(2850)(7.8\cdot 10^{17})=2.2\cdot 10^{21} kg[/tex]

2)

The kinetic energy of the continent is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

[tex]m=2.2\cdot 10^{21} kg[/tex] is its mass

v = 3.8 cm/year is its speed

We have to convert the speed into m/s. Keeping in mind that

1 cm = 0.01 m

[tex]1 year = 365\cdot 24\cdot 60 \cdot 60 =3.15\cdot 10^7 s[/tex]

We find

[tex]v=3.18 \frac{cm}{y} \cdot \frac{0.01}{365\cdot 24 \cdot 60 \cdot 60}=1.0\cdot 10^{-9} m/s[/tex]

So now we can find the kinetic energy:

[tex]K=\frac{1}{2}(2.2\cdot 10^{21})(1.0\cdot 10^{-9})^2=1118 J[/tex]

3)

The kinetic energy of the jogger is given by

[tex]K=\frac{1}{2}m'v'^2[/tex]

where

m' = 80 kg is the mass of the jogger

v' is the speed of the jogger

Here we want the jogger to have the same kinetic energy of the continent, so

[tex]K=1118 J[/tex]

And by re-arranging the equation, we can find what speed the jogger must have:

[tex]v'=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1118)}{80}}=5.3 m/s[/tex]

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Answer:

SOUND ENERGY is the energy which gets released or formed.

Explanation:

Remember , Energy is ALWAYS conserved. It can neither be destroyed nor be created(made).

∴

Energy gets converted significantly to sound energy - from the following options.

Final answer:

The type of energy made by a door slamming is sound energy, which is a result of the door's mechanical energy being partially transformed into pressure waves in the air.

Explanation:

When a door slams, it produces sound energy as a result of the forceful closure that creates pressure waves in the air we perceive as sound. While the door itself experiences mechanical energy, which is necessary for the motion of slamming, the type of energy being asked about in this case is the one that is made as a consequence of that action, which is sound energy. Mechanical energy is the sum of kinetic and potential energy within a system and is responsible for physical movement, as when a person pushes a door causing it to slam. Upon impact, part of this mechanical energy is transformed into sound energy.

Answer:

B

Explanation:

F = ma , a = F/m

a1 = F/10 and a2 = F/4

Since Force is constant, a2 will we greater than a1

Answer:

Answer: C

explanation:

They could be same or different

ie:

|-5|,|-5| = 5,5

|-5|,|5| = 5,5

Answer:

C

Explanation:

9.8 m/s/s as well

Answer:

look at the explanation

Explanation:

This pith-ball electroscope is used to detect the presence of a static electricity charge. The two lightweight “pith” balls suspended from the strings are attracted to objects with a static electric charge. The pith balls can also be charged by touching them to an object with a static electric charge.

Answer:

B)velocity of object decreases

Explanation:

Consider the positive x axis as positive direction

Assume a body moving in negative x-axis direction

It's acceleration also alone negative x-axis direction

So according to our consideration

velocity and acceleration values are negative

That is both are towards negative x direction

But as both velocity and acceleration are in same direction, MAGNITUDE of velocity increases

But as magnitude increases in negative direction, velocity value decreases

But speed value increases(As speed is scalar and velocity is a vector)

Answer:

(C) Plants use oxygen to make their own food

Answer:

B.

Explanation:

We can use process of elimination here.  A is not correct.  Animals do not make their own food.  C is not correct.  Plants use carbon dioxide, not oxygen to power photosynthesis.  D is not correct.  Decomposers break down dead animals.  Therefore, B is correct.  Plants are food for some consumers.  

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

[tex]\mu mg = m\frac{v^2}{r}[/tex]

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

[tex]\mu=0.40[/tex] is the coefficient of friction between the tires and the road

m is the mass of the car

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

[tex]v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s[/tex]

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

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Final answer:

On a rainy day, the maximum speed at which a curve with a radius of 150 m can be safely negotiated is 12.43 m/s. This calculation is based on the centripetal force provided by the friction between the tires and the road surface.

Explanation:

To determine the maximum speed at which the curve can be safely negotiated, we need to consider the centripetal force acting on the car. The centripetal force is provided by the friction between the tires and the road surface. The formula for centripetal force is Fc = mv²/r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.

On a rainy day, the coefficient of friction between the tires and the road is given as 0.40. The maximum frictional force is given by the equation Ff = μN, where μ is the coefficient of friction and N is the normal force.

Since the normal force is equal to the gravitational force acting on the car, we can calculate it using N = mg, where m is the mass of the car and g is the acceleration due to gravity.

To find the maximum speed, we need to equate the centripetal force and the frictional force: mv²/r = μN. We can rearrange this equation to solve for the maximum velocity v: v = sqrt(μrg).

Substituting the given values, v = sqrt(0.40 * 9.8 * 150) = 12.43 m/s.

1) Frequency: 1.95 Hz

2) Equilibrium position: 0.138 m

3) Amplitude: 0.357 m

4) Maximum speed: 4.36 m/s

5) Maximum acceleration: [tex]53.1 m/s^2[/tex]

6) Spring constant: 15.0 N/m

7) Total mechanical energy: 0.96 J

Explanation:

1)

The frequency of a simple harmonic motion is equal to the reciprocal of the period:

[tex]f=\frac{1}{T}[/tex]

where

f is the frequency

T is the period

For the particle in this problem, we have

T = 0.513 s (period)

So the frequency of motion is

[tex]f=\frac{1}{0.513}=1.95 Hz[/tex]

2)

In a simple harmonic motion, the object oscillates between two maximum positions [tex]+x_m[/tex] and [tex]-x_m[/tex] which are equidistance from the equilibrium position. So, the equilibrium position is the midpoint between these two positions.

For the particle in this problem, the two extreme positions are:

[tex]x_1 = -0.219 m[/tex]

[tex]x_2 = 0.495 m[/tex]

So the mid-point (the equilibrium position) is

[tex]x_{ep} = \frac{x_1 + x_2}{2}=\frac{-0.219+0.495}{2}=0.138 m[/tex]

3)

The amplitude of a simple harmonic motion is the maximum displacement of the object, measured from the equilibrium position.

This means that we can calculate the amplitude simply as the difference between one of the extreme positions and the equilibrium position.

Taking

[tex]x_2 = 0.495 m[/tex]

and

[tex]x_{ep} = 0.138 m[/tex]

We find the amplitude:

[tex]A=x_2 - x_{ep} = 0.495-0.138 =0.357 m[/tex]

4)

In a simple harmonic motion, the maximum speed is given by

[tex]v_{max}=\omega A[/tex]

where

[tex]\omega[/tex] is the angular frequency

A is the amplitude

The angular speed can be calculated from the frequency as follows:

[tex]\omega=2\pi f=2 \pi (1.95 Hz)=12.2 rad/s[/tex]

The amplitude is

A = 0.357 m

So, the maximum speed is

[tex]v_{max} = (12.2)(0.357)=4.36 m/s[/tex]

5)

The maximum acceleration in a simple harmonic motion is given by

[tex]a_{max}= \omega^2 A[/tex]

Where we already know that:

[tex]\omega=12.2 rad/s[/tex] is the angular frequency

A = 0.357 m is the amplitude of motion

Substituting, we find the maximum acceleration:

[tex]a_{max}=(12.2)^2(0.357)=53.1 m/s^2[/tex]

6)

The angular speed in a simple harmonic motion can be written as

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here we know that:

[tex]\omega=12.2 rad/s[/tex] is the angular speed

m = 0.101 kg is the mass of the particle

So we can solve the formula for k, the spring constant:

[tex]k=\omega^2 m =(12.2)^2(0.101)=15.0 N/m[/tex]

7)

Since the energy is conserved, the total mechanical mechanical energy is just equal to the maximum potential energy of the system, which occurs when the particle is at maximum displacement (x=A) and the speed is zero (so the kinetic energy is zero), therefore it is given by:

[tex]E=\frac{1}{2}kA^2[/tex]

where

k = 15.0 N/m is the spring constant

A = 0.357 m is the amplitude

And substituting,

[tex]E=\frac{1}{2}(15.0)(0.357)^2=0.96 J[/tex]

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Answer:

The mass of the earth is, M = 5.98 x 10²⁴ kg

Explanation:

Given data,

The acceleration due to gravity, g = 9.8 m/s²

The radius of Earth, R = 6400 km

                                    =6.4 x 10⁶ m

The universal gravitational constant, G = 6.67 x 10 ⁻¹¹ Nm²kg⁻²

Using the equation,

                         GM = gR²

                            M = gR² / G

Substituting the values,

                            M = 9.8 x (6.4 x 10⁶)² / 6.67 x 10 ⁻¹¹

                                = 5.98 x 10²⁴ kg

Hence, the mass of the earth is, M = 5.98 x 10²⁴ kg

The force of friction is 25 N

Explanation:

For this problem, we can apply Newton's second law of motion along the horizontal direction:

[tex]\sum F_x = ma_x[/tex]

where

[tex]\sum F_x[/tex] is the net force in the horizontal direction

m is the mass of the block

[tex]a_x[/tex] is the horizontal acceleration

Here  the block is moving at constant speed, so its acceleration is zero, therefore:

[tex]a=0 \rightarrow \sum F_x = 0[/tex] (1)

The net force in the horizontal direction can be written as:

[tex]\sum F_x = Fcos \theta -F_f[/tex] (2)

where

Combining (1) and (2), we find the magnitude of the force of friction:

[tex]Fcos \theta -F_f = 0\\F_f = F cos \theta =(50)(cos 60^{\circ})=25 N[/tex]

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Answer:

A

Explanation:

Answer:

a

Explanation:

a