Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05 m/s2. Using Newton's Second Law, you can compute how much force Mike is applying to the car.

Answer:

0.146 m/s

Explanation:

We can see it in the pic.

Answer:

√2µgd = vi

Explanation:

Given that the mass of object is m

Sliding speed of object is vi

coefficient of kinetic friction is µ

Distance covered is d

Final speed, vf is 0

Equation to apply will be;

v²f=v²i-2µgd where  g is acceleration due to gravity

v²f+2µgd=v²i----------------------but v²f = 0

2µgd=v²i

√2µgd = vi

In the physical scenario described, the most reasonable formula for calculating the object's initial speed across a level tabletop, given the coefficient of kinetic friction, the object's mass, and the distance moved, would be vi is equal to the square root of twice the coefficient of friction, the mass of the object, gravity, and the distance traveled.

The object in question moves with initial speed vi across a level tabletop under the influence of kinetic friction, and eventually comes to a stop after moving through a distance of d. According to the principles of physics, specifically those pertaining to motion and kinetic friction, the initial speed of an object can be inferred from an equation that takes into account the weight of the object (mass m times gravity g), the distance traveled d, and the coefficient of kinetic friction µ. Considering this, the equation which makes the most sense in this situation would be vi = √2µmgd, where the initial speed vi is equal to the square root of twice the coefficient of friction µ times the mass of the object m, gravity g, and the distance d.

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Answer:

[tex]F_n = 185.72 N[/tex]

Explanation:

As per FBD we can say that net downward force perpendicular to contact plane must be counterbalanced by the net upward Normal force

So here we will say

[tex]F_n = F sin40 + mg[/tex]

now we have

[tex]F = 60.0 N[/tex]

[tex]m = 15 kg[/tex]

now we have

[tex]F_n = 60 sin40 + 15(9.81)[/tex]

[tex]F_n = 185.72 N[/tex]

The normal force that the floor exerts on the chair is approximately 185.72 N.

First, let's draw a free-body diagram for the chair.

The forces acting on the chair are the force you apply (60.0 N) and the weight of the chair (mg), where m is the mass of the chair (15.0 kg) and g is the acceleration due to gravity (9.81 m/s^2).

Now, since the chair is sliding along the floor, there must be a frictional force opposing its motion. However, the problem doesn't provide the coefficient of friction or any information about it. Therefore, we cannot consider the frictional force in this calculation.

Next, let's break down the force you apply into its components. The force you apply is directed at an angle of 40.0° below the horizontal. We can resolve this force into two components: one perpendicular to the floor and one parallel to the floor.

The component perpendicular to the floor is given by F * sinθ, where F is the magnitude of the force (60.0 N) and θ is the angle (40.0°). This component contributes to the normal force.

The component parallel to the floor is given by F * cosθ. This component contributes to the frictional force, but since we are not considering friction in this calculation, we won't use this component.

Now, let's calculate the normal force. The normal force is the net force acting perpendicular to the contact plane. We can find it by adding the force component perpendicular to the floor (F * sinθ) and the weight of the chair (mg).

So, the normal force (Fn) is given by Fn = F * sinθ + mg.

Substituting the given values, we have:

Fn = 60.0 N * sin(40.0°) + 15.0 kg * 9.81 m/s².

Evaluating this expression, we find that the normal force is approximately 185.72 N.

Therefore, the normal force that the floor exerts on the chair is approximately 185.72 N.

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The trombone becomes out of tune when moving from indoors to the field because of the temperature change affecting the air density and speed of sound within the instrument's tubing, thus altering the pitch.

When the marching band member tunes their trombone indoors and then finds it out of tune upon moving to the field, the primary cause is likely due to a change in temperature. Brass instruments, like trombones, are sensitive to temperature changes because they directly affect the air's density inside the instrument's tubing. Warmer air makes the instrument sound sharper, while colder air will cause it to sound flatter. Since the trombone changes its pitch by varying the length of the tube through its slide mechanism, a change in outdoor temperature can significantly alter its tuning. This is because the warmer or colder air affects the speed of sound in the air column, changing the resonant frequencies of the tube and thus the pitch produced.

Answer:

The answer to your question is:  magnitude:  1534.9 units

direction: 326.5°

Explanation:

Data

x = 1280 u

y = -847 y

Then as x is positive and y is negative this vector is quadrangle

To find the magnitude, we use the pythagorean theorem

c2 = a2 + b2

c2 = 1280² + (-847)² = 1638400 + 717409 = 2355809

c = 1534.9 units

to find the direction we use the tangent function

tanФ = os/as = -847/1280 = -0.661

Ф = 33.49

But it is in the 4th quadrangle, then

Ф = 360 - 33-49 = 326.5°

Answer:

1530 units, -33.5°

Explanation:

Given the x-component and the y-component, the magnitude can be found with Pythagorean theorem:

v² = vx² + vy²

And the direction can be found with trigonometry:

θ = atan(vy / vx)

Given that vx = 1280 and vy = -847:

v² = (1280)² + (-847)²

v ≈ 1530

θ = atan(-847 / 1280)

θ ≈ -33.5° or 146.5°

θ is in the fourth quadrant, so θ = -33.5°.

The unknown frequency of the other whistle (brand x) is either 28.5 kHz or 18.5 kHz. This is because when both are played together, they create a beat frequency of 5000 Hz due to wave interference.

The perception of frequency is called pitch. Humans, typically, can distinguish between two sounds if their frequencies differ by roughly 0.3%, but they cannot hear frequencies beyond the 20,000 Hz range. If a 5000 Hz frequency is created when the two dog whistles are played together, it suggests that some form of wave interference is occurring. This particularly seems like a case of beat frequency, which is the resultant frequency produced due to the superposition of two sound waves of different frequencies.

Given that one whistle operates at 23.5 kHz (or 23500 Hz), and the beat frequency when both whistles are played together is 5000Hz, the other whistle (brand x) likely operates at either 23500 + 5000 = 28.5 kHz (or 28500 Hz) or 23500 - 5000 = 18.5 kHz (or 18500 Hz). Either of these frequencies would yield a 5000 Hz difference when superimposed with the 23500 Hz frequency, thereby creating a beat frequency of 5000 Hz when the two are played together.

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The situation described is a phenomenon called beat frequency. Since humans cannot hear the individual sounds of the whistles which must be above 20kHz but only a beat frequency of 5000 Hz, the frequency of the brand X whistle is calculated to be 28.5 kHz.

The scenario described in this situation is an example of beat frequency, which is a phenomenon in physics. When two waves of slightly different frequencies interfere with each other, they produce a beat frequency which is the difference of the initial two frequencies. Considering the provided information, we know a shrill whine of frequency 5000 Hz is heard when the two dog whistles are played together. One whistle operates at a frequency of 23.5 kHz (or 23500 Hz).

Since the beat frequency is the difference between the two initial frequencies, the frequency of the whistle brand X (denoted as f) can be calculated by subtracting or adding the beat frequency from the known whistle frequency. This means that the unknown frequency could be either 23500 Hz - 5000 Hz = 18500 Hz, or 23500 Hz + 5000 Hz = 28500 Hz. However, since the upper limit of human hearing is 20 kHz and humans can't hear the shrill whine produced by either whistle individually, the frequency of brand X must be higher than 20 kHz and therefore is 28500 Hz.

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Answer:

given,

Probability of student studying math P(M)=[tex]\dfrac{45}{100} = 0.45[/tex]

Probability of student studying physicsP(P) = [tex]\dfrac{61}{100} = 0.61[/tex]

Probability of student studying both math and physics together P(M∩P) = [tex]\dfrac{25}{100} = 0.25[/tex]

a) student took mathematics or physics

P(M∪P) = P(M) + P(P) - P(M∩P)

             = 0.45 + 0.61  - 0.25

             = 0.81

b) student did not take either of the subject

P((M∪P)') = 1 - 0.81

               = 0.19

c) Student take physics but not mathematics

P(P∩M') = P(P) - P(P∩M)

             = 0.61 - 0.25

             = 0.36

studying physics and mathematics is  not mutually exclusive because we can study both the subjects.

The probability that a student took mathematics or physics is 83%, did not take either is 17%, and took physics but not mathematics is 36%. Mathematics and physics are not mutually exclusive since some students studied both.

The problem describes a situation where students in a high school graduating class have taken either mathematics, physics, or both. The task is to determine various probabilities based on this information. The solution requires an understanding of basic set theory and probability concepts.

Probability Calculations

Mathematics and physics are not mutually exclusive events because 25 students studied both. If they were mutually exclusive, no students would have studied both subjects.

Answer:

The plane will be 4795.23 meters from where he threw the package.

Explanation:

Data:

v = 167 m/s

h = 4040 m

g = 9.8m/s²

package:

after it is dropped there is no horizontal force, so...

[tex]X - Xo = Vt[/tex]

t = [tex]\frac{X - Xo}{V}[/tex]  I

in the vertical we have only [tex]P = mg[/tex]

[tex]Y - Yo = Vot -\frac{gt^{2} }{2}[/tex] but Vo = 0 because the package is dropped

[tex]Y - Yo =-\frac{gt^{2} }{2}[/tex]  II

replacing I in II

[tex]Y - Yo =-\frac{g(X - Xo)^{2} }{2V^{2}}[/tex]

[tex]0 - 4040 =-\frac{9.8(X - Xo)^{2} }{2*167^{2}}[/tex]

X - Xo = 4795.23

The distance from where the plane drops the package and where it hits the ground is the same as the plane flies horizontally, as there is no acceleration at x.

Answer:

97.03%

Explanation:

The equation for volumetric expansion due to thermal expansion is as follows

V/Vo=(1+γΔT)

V=final volume

Vo=initial volume

γ=coefficient of volume expansion=3.2 × 10–5 K–1

ΔT=

temperature difference

assuming that the earth is a sphere the volume is given by

V=(4/3)pi R^3

if we find the relationship between the initial and final volume we have the following

[tex]\frac{V}{Vo}  =\frac{ \frac{4}{3} \pi r^{3} }{ \frac{4}{3} \pi ro^{3}}=\frac{r^{3} }{ro^{3}}[/tex]

taking into account the previous equation

r/ro=(1+γΔT)^(1/3)

r/r0=(1-3.2x10-5(3000-300))^(1/3)=

r/ro=0.9703=97.03%

Answer:

127.36 m/s

Explanation:

velocity of plane with respect to air = 140 m/s due east

velocity of air with respect to ground = 31 m/s 30° west of north

Write the velocities in the vector forms

[tex]\overrightarrow{V_{p/a}}=140\widehat{i}[/tex]

[tex]\overrightarrow{V_{a/g}}=31  \left ( -Sin30 \widehat{i}+Cos30\widehat{j} \right )[/tex]

[tex]\overrightarrow{V_{a/g}}= -15.5 \widehat{i}+26.85\widehat{j}[/tex]

Let velocity of plane with respect to ground is given by vp/g

According to the formula of relative velocities

[tex]\overrightarrow{V_{p/a}}=\overrightarrow{V_{p/g}}-\overrightarrow{V_{a/g}}[/tex]

[tex]\overrightarrow{V_{p/g}}=\overrightarrow{V_{p/a}}+\overrightarrow{V_{a/g}}[/tex]

[tex]\overrightarrow{V_{p/g}}= \left ( 140-15.5 \right )\widehat{i}+26.85\widehat{j}[/tex]

[tex]\overrightarrow{V_{p/g}}= \left ( 124.5 \right )\widehat{i}+26.85\widehat{j}[/tex]

The magnitude of the velocity of plane with respect to the ground is given by

[tex]V_{p/g} = \sqrt{124.5^{2}+26.85^{2}}=127.36 m/s[/tex]

Thus, the velocity of plane with respect to the ground is given by 127.36 m/s.

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

The branch must provide an upward force of 227 Newtons to support a gibbon weighing 8.5 kg, swinging with a speed of 3.2 m/s, and a radius of 0.6 m.

To calculate the upward force that the branch must provide, we need to take into account both the gravitational force acting on the gibbon and the centrifugal force due to its circular motion. The weight of the gibbon (gravitational force) is its mass times the acceleration due to gravity (8.5 kg * 9.8 m/s^2), which equals 83.3 N.

Centrifugal force is given by mass times velocity squared divided by radius of the motion (m*v^2/r), which results in (8.5 kg * (3.2 m/s)^2) / 0.6 m = 144 N.

By adding these two forces together, we find the total force the branch must supply at the lowest point: 83.3 N + 144 N = 227 N.The branch must provide an upward force of 227 Newtons to support the swinging gibbon.

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Explanation:

When an object is moving in a circular path, the motion of the object is called uniform circular motion. The object moves under the action of centripetal acceleration. It is given by :

[tex]a=\dfrac{v^2}{r}[/tex]

r is the radius of circular path

v is the speed of the object

In uniform circular motion, the object moves with constant speed. Also, the velocity of the object keeps on changing because it changes direction at every instant of time. Also, the object is accelerating due to change in velocity.

So, the correct options are (b) and (c).

In uniform circular motion, an object travels in a circular path at a constant speed, but its velocity is not constant because the direction of motion changes. Thus, the object is accelerating due to the continuous change in direction, despite the speed being constant.

Understanding Uniform Circular Motion

Uniform circular motion refers to the motion of an object traveling in a circular path at a constant speed. This situation presents a peculiar form of acceleration. Even though the object maintains a constant speed, its velocity is not constant because its direction is continually changing.

The three statements provided, when evaluated, give us the following insights:

Therefore, the correct statements about an object in uniform circular motion are that it moves at a constant speed and it is indeed accelerating.

Answer:

Option D.

Explanation:

The first law of thermodynamics is a law of conservation of energy. This automatically tells us that energy lost by a system won't dissapear, but it will be gained by the surroundings.

Mathematically is stated this way:

ΔU=Q-W

Where ΔU is the change in the internal energy of a closed system, Q is the amount of heat supplied and W the amount of work done by the system on its surroundings. It means that if the internal energy U decreases, then that energy lost must have been converted to work W in the surroundings.

The first law of thermodynamics is best represented by the statement: Energy lost by the system must be gained by the surroundings.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed but it can be transformed from one form to another. The most accurate representation of the first law of thermodynamics from the provided options is [D] Energy lost by the system must be gained by the surroundings. This statement effectively communicates the balance of energy transfer which is fundamental to understanding the concept of first law of thermodynamics.

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Answer:

[tex]H = \frac{u^2}{2g} + h[/tex]

Explanation:

Let the football is kicked up vertically with some speed given as

[tex]v = v_o[/tex]

now its speed when it will reach to height "h" above the ground is given as "u"

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we have

[tex]u^2 - v_o^2 = 2(-g)h[/tex]

so we have

[tex]v_o^2 = u^2 + 2gh[/tex]

now we know that when football will reach to maximum height then it will have zero final velocity

So we will have

[tex]v_f^2 - v_i^2 = 2 a s[/tex]

[tex]0 - (u^2 + 2gh) = 2(-g)H[/tex]

so we have maximum height given as

[tex]H = \frac{u^2 + 2gh}{2g}[/tex]

[tex]H = \frac{u^2}{2g} + h[/tex]

The maximum height a football reaches after being kicked upward past a window can be computed by using the kinematic equation and setting the final velocity to zero at the peak. The formula H = h + (U²) / (2g) allows us to find the maximum height by adding the window's height above ground to the kinetic energy at the window level, divided by twice the acceleration due to gravity.

Calculating the Maximum Height of a Football Kicked Upward

To find the maximum height a football reaches after being kicked upward past a window at velocity U and height h, we can use the principles of kinematics under constant acceleration due to gravity. Initially, when the ball is kicked, it possesses kinetic energy which gets converted into gravitational potential energy as it ascends. The football reaches its maximum height when its velocity becomes zero.

Using the kinematic equation:

v² = u² + 2gh

Where:

Let H be the maximum height above ground and h the window height above ground, then:

0 = U² - 2g(H - h)

Solving for H gives us:

H = h + (U²) / (2g)

Plugging in the values for U and h as well as g, we can calculate the maximum height H.

If the airplane is carrying 58 gallons of fuel, it will weigh 2452 pounds.

Given that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank.

Carrying 18 gallons of fuel, the airplane weighs 1999 pounds.

Also carrying 46 gallons of fuel, it weighs 2153 pounds.

First we need to find the difference in gallons and the difference in the weights

46 - 12 = 34 gallons

2399 - 2176 = 223 pounds

Therefore, 34 gallons of gas weighs 223 pounds

Now the weight per pound:

223 pounds / 34 gallons = 6.5588 pounds per gallon

Similarly,

54 gallons - 46 gallons = 8 gallons

8 gallons x 6.6 pounds per gallon = 52.8 pounds, round to 53 pounds

Now Total weight of plane with 54 gallons

2399 + 53 = 2452

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The weight of the airplane is a linear function of the amount of fuel in its tank. The slope of this function is found to be 5.5 pounds per gallon of fuel. Substituting into the equation, we find that the airplane weighs 2217 pounds when it is carrying 58 gallons of fuel.

This question can be solved by using the concept of linear function in mathematics. The weight of the airplane changes linearly with the amount of fuel in its tank. This gives us two points on the linear function (18, 1999) and (46, 2153). We can find the slope of this function by subtracting the y coordinates and dividing by the difference of the x coordinates, which gives us (2153-1999)/(46-18)=154/28 = 5.5 pounds/gallon. This is the amount the airplane's weight changes for each gallon of fuel. To find the weight of the airplane when it is carrying 58 gallons of fuel, we can use the point-slope form of a line, y-y1=m(x-x1), using one of our points and the slope we found. Plugging in gives us Weight - 1999 = 5.5×(Fuel - 18), which simplifies to Weight = 5.5*Fuel + 1999 - 5.5×18. Substituting 58 in place of Fuel, we get a weight of 2217 pounds.

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-- Gravity slows the ball by 9.8 m/s every second. So it slows to zero and reached its highest point in

(58.8 / 9.8) = 6 seconds .

-- Its average speed all the way up is

(1/2) (58.8 + 0) = 29.4 m/s

-- Traveling for 6 seconds at an average speed of 29.4 m/s, the ball covers

(6) (29.4) = 176.4 meters .

Final answer:

Robbie Knievel's launch angle for his jump across the Grand Canyon was approximately -5.35 degrees.

Explanation:

To calculate the launch angle of Robbie Knievel's jump across the Grand Canyon, we can use the horizontal distance and initial velocity of the motorcycle. The horizontal distance is given as the width of the canyon, which is 65 meters. The initial velocity is given as 36 m/s. We can use the equation for horizontal motion to find the time of flight:

d = v0xt

Where d is the horizontal distance, v0x is the initial velocity in the x-direction, and t is the time of flight. Solving for t, we get:

t = d / v0x

Substituting the given values, we have:

t = 65 m / 36 m/s = 1.81 s

Next, we can use the equation for vertical motion to find the launch angle:

y = v0yt + (1/2)gt2

Where y is the vertical distance, v0y is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time of flight. The vertical distance is zero since Robbie reached the other side of the canyon. Solving for v0y and using the given value of t, we get:

v0y = - (1/2)gt

v0y = - (1/2)(3.7 m/s2)(1.81 s) = -3.34 m/s

Finally, we can use the launch angle formula:

tan(θ) = v0y / v0x

Substituting the calculated values, we have:

tan(θ) = -3.34 m/s / 36 m/s

θ = atan(-3.34 m/s / 36 m/s) = -5.35 degrees

Therefore, Robbie Knievel's launch angle was approximately -5.35 degrees.

Answer:

13.7 s

Explanation:

The position at time t of car A can be written as follows:

[tex]x_A (t) = \frac{1}{2}a_At^2[/tex]

where

[tex]a_A = 11 m/s^2[/tex] is the acceleration of car A

The position of car B instead can be written as

[tex]x_B(t) = d+\frac{1}{2}a_B t^2[/tex]

where

[tex]a_B = -4 m/s^2[/tex] is the acceleration of car B

d = 1400 m is the initial separation between the cars

The two cars meet when

[tex]x_A = x_B[/tex]

Using the two equations above,

[tex]\frac{1}{2}a_A t^2 = d + \frac{1}{2}a_B t^2\\\frac{1}{2}t^2 (a_A - a_B) = d\\t=\sqrt{\frac{2d}{a_A-a_B}}=\sqrt{\frac{2(1400)}{11-(-4)}}=13.7 s[/tex]

Final answer:

The two cars with given accelerations and an initial distance of 1400 meters between them when starting from rest will meet after 20 seconds.

Explanation:

To solve for the time at which the two cars meet, we must consider the accelerations of both cars and the initial distance between them. Assuming car A is moving with a positive acceleration of 11 m/s2 and car B is moving with a negative acceleration of -4 m/s², we need to find a common point in time where they both cover the total distance of 1400 m when starting from rest.

Let the time taken for the cars to meet be denoted by 't'. For car A, the displacement (sA) is given by the formula sA = 0.5 * aA * t², and for car B, the displacement (sB) is given by sB = 0.5 * aB * t². As they are moving towards each other, the sum of their displacements sA+sB should equal the initial separation distance which is 1400m.

This gives us the equation 0.5 * 11 * t2 + 0.5 * (-4) * t² = 1400. Simplifying, we get 3.5 * t² = 1400, and solving for 't' gives us t2 = 400, so t = 20 seconds. Hence, both cars will meet after 20 seconds.

Answer:

(a) 52.724 m/s

(b) Total displacement, d = 551.25 m

Solution:

As per the question:

Initial acceleration of the speed boat, a = - 2.01 [tex]m/s^{2}[/tex]

Time duration, t = 7.00 s

Additional time, t' =6.00 s

Acceleration for additional time, a' = 0.518 [tex]m/s^{2}[/tex]

The followed up acceleration, a'' = 1.49 [tex]m/s^{2}[/tex]

Time duration, t'' = 8.00 s

(a) Now, to calculate the velocity of the boat at timer, t = 21.0 s, we have:

After the initial 7.00 s, the velocity of the boat, from eqn-1 of motion:

v = u + at

v = 0 - 2.1(7.00) = - 14.7 m/s

After t + t' = 13 s:

v' = v + at

v' =  14.7 + 0.518(13) =  21.434 m/s

Now, velocity of the boat after t = 21 s:

v'' = v' + a''t

v'' = 21.434 + 1.49(21) = 52.724 m/s

(b) Now, the total displacement, d:

For the first case:

d = ut + [tex]\frac{1}{2}at^{2} = 0 - 0.5\times 2.1\times 7^{2} = - 51.45 m[/tex]

For the second case:

d = v't' = 21.434(6) = 128.6 m

For the third case:

d = ut + [tex]\frac{1}{2}a't'^{2} = 0 + 0.5\times 0.518\times 6^{2} = 4.65 m[/tex]

For the fourth case:

d = v''t'' = 52.724(8) = 421.79 m

For the last case:

d = ut + [tex]\frac{1}{2}a't'^{2} = 0 + 0.5\times 1.49\times 8^{2} = 47.68 m[/tex]

Total displacement, d = -51.45 + 128.6 + 4.65 + 421.79 + 47.68 = 551.25 m

Answer:

There is an attractive force between the rod and sphere.

Explanation:

When negatively charged rod is placed close to the metal sphere then due to the electric field of the rod the opposite free charge of metal sphere comes closer to the rod on one surface

While similar charge in the metal sphere move away from the rod due to repulsion of electric field of rod

This temporary charge distribution of the metal sphere is known as induction

And since opposite charge on the metal surface comes closer to the metal sphere so here we can say that the rod will attract the metal sphere

so here correct answer will be

There is an attractive force between the rod and sphere.

Answer:

There is an attractive force between the rod and sphere.

Explanation:

Answer:

c. compound light microscope

Explanation:

Microorganisms (such as bacteria, fungi and protozoa) can be observed through a compound light microscope, which allows to increase the observed image since 400 times to 1000 times (in more advanced microscopes). This kind of organisms have sizes in the order of micrometers (μm, a million times lesser than a meter), so you cannot observe them with a dissecting microscope, which only increase the image 10-30 times. In fact, dissecting microscopes are employed to observe thin sections of tissues (of plants and animals) and, as the name say it, to "dissect" in very tiny parts.

The other two microscopes, transmission electron and scanning electron microscopes, are used to observe macromolecules, and physical changes in compounds and matter, in the order of nanometers (nm, a thousand million times lesser than a meter).

Final answer:

The scientist should use a compound light microscope to observe and draw living microorganisms from pond water, as it allows for the necessary magnification while preserving the organisms' life for movement study.

Explanation:

To observe and draw the structure of microorganisms in pond water and study their movement, the best type of microscope for a scientist to use would be a compound light microscope. This type of microscope uses visible light that passes through and is bent by the lens system, allowing the user to observe living organisms, which is essential for studying movement. Moreover, light microscopes can magnify cells up to approximately 400 times, which is typically sufficient for viewing microorganisms.

While electron microscopes like the transmission electron microscope and scanning electron microscope provide much higher magnification and resolution, they are not suitable for observing living specimens because the sample preparation process kills the organisms. Dissecting microscopes, on the other hand, provide a three-dimensional view of the specimen but have lower magnification and are more suitable for larger objects such as tissues, not microorganisms in pond water.

To find the time it takes for the baseball to reach the ground, we can use the equations of motion for an object in free fall. We can solve a quadratic equation to find the time and then use it to find the final velocity of the baseball when it strikes the ground.

To solve this problem, we can use the equations of motion for an object in free fall. The object is thrown downward, so the initial velocity is negative. We can use the equation: h = ut + (1/2)gt² to find the time taken to reach the ground. In this equation, h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity. Since the initial height is 555 ft and the initial velocity is 40 ft/s, the equation becomes 555 = -40t + (1/2)(32)(t²). We can solve this quadratic equation to find the time it takes for the baseball to reach the ground. Once we have the time, we can use the equation: v = u + gt to find the final velocity. Plugging in the values, we get v = 40 + (32)t. Substitute the value of t from the first equation to find the final velocity when the baseball strikes the ground.

Answer:

The net force acting on an object is the sum of all the force acting on it, and the net force of an object is zero. I f the forces acting on it tend to cancel each other. For example you are sit in a chair, the earth's gravity is pulling you down, but the chair is pushing you up with an equal amount of force.

Explanation:

Answer:

The net force acting on an object is the sum of all the force acting on it, and the net force of an object is zero. I f the forces acting on it tend to cancel each other. For example you are sit in a chair, the earth's gravity is pulling you down, but the chair is pushing you up with an equal amount of force.

Explanation:

science teacher helped me

Answer:

[tex]T = 175.6 N[/tex]

Explanation:

As we know that string is vibrating in third harmonic

So we will have

[tex]L = 3\frac{\lambda}{2}[/tex]

so we have

[tex]0.79 = \frac{3}{2}\lambda[/tex]

so we have

[tex]\lambda = \frac{2}{3}(0.79)[/tex]

[tex]\lambda = 0.527[/tex]

we know that frequency of the wave is given as

f = 500 Hz

now we know that

speed of the wave is

[tex]v = frequency \times wavelength[/tex]

[tex]v = (500)(0.527)[/tex]

[tex]v = 263.3 m/s[/tex]

now we have

[tex]v = \sqrt{\frac{T}{m/L}}[/tex]

so we have

[tex]263.3 \sqrt{\frac{T}{(0.002/0.79)}}[/tex]

[tex]T = 175.6 N[/tex]

Answer:

The lifeguard takes 25.9  seconds to reach the child, at 25.9 meters from the start point downstream.

Explanation:

As the image shows, the child trajectory, the lifeguard trajectory and the distance from the bank form a triangle. This triangle is formed by the distances, an we already know the distance from the bank and the speed of child, and the speed of the lifeguard. So we have unknom time in common. Lets see the equations:

Using phitagoras theorem

[tex]45^{2}+(1*t_{1} )^{2}  =(2*t_{2} )^{2}\\\\but\\t_{1} =t_{2} , then\\\\45^{2} =3t^{2} \\\\t=\sqrt{\frac{45^{2}}{3}  } = 25.9seconds\\and replacing in X1= 25.9 meters[/tex]

Final answer:

The lifeguard takes 22.5 seconds to reach the child, intercepting 22.5 meters downstream due to the river's current. This solution involves calculating relative speeds and distances, showcasing principles of physics.

Explanation:

The question describes a scenario where a lifeguard needs to save a child being carried downstream by a river's current. The lifeguard can swim at a speed of 2.0 m/s relative to the water, and the child is 45 meters away from the riverbank, being carried by a current of 1.0 m/s. To solve for the time it takes for the lifeguard to reach the child and the distance downstream where the interception occurs, we must analyze the relative velocities and distances involved.

Step 1: Determine the Time to Reach the Child

The lifeguard's effective speed towards the child in the direction perpendicular to the current is 2.0 m/s. Since the child is 45 m from the bank and the lifeguard swims directly towards the child, the time it takes will be Time = Distance / Speed = 45 m / 2.0 m/s = 22.5 seconds.

Step 2: Determine the Distance Downstream

During this time, both the child and the lifeguard are being carried downstream by the current. The distance covered downstream can be calculated by the current's speed multiplied by the time: Distance downstream = Current speed × Time = 1.0 m/s × 22.5 s = 22.5 meters.

Therefore, it takes the lifeguard 22.5 seconds to reach the child, and they intercept 22.5 meters downstream from the point directly opposite the lifeguard's starting position.

Answer:

The distance moved is 9 meters

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Explanation:

- Distance is the length of the actual path between the initial and the

  final position. Distance is a scalar quantity

- Displacement is the change in position, measuring from its starting

  position to the final position. Displacement is a vector quantity

The quarterback pedals 3.3 meters southward

That means it moves down 3.3 meters

Then runs 5.7 meters northward

That means it runs up 5.7 meters

The distance = 3.3 + 5.7 = 9 meters

The distance moved is 9 meters

It moves southward (down) for 3.3 meters and then moves northward

(up) for 5.7

It moves from zero to 3.3 down and then moves up to 5.7

The displacement = 5.7 - 3.3 = 2.4 meters northward

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Final answer:

The quarterback moved a total distance of 9.0 meters. The displacement of the quarterback was 2.4 meters northward, as displacement takes into account the direction of motion.

Explanation:

When considering the movement of a quarterback who backpedals 3.3 meters southward and then runs 5.7 meters northward, we need to determine both the total distance moved and the magnitude and direction of the displacement.

The distance is a scalar quantity that represents the total path length traveled, regardless of direction. In this case, the quarterback moved a total distance of 3.3 meters + 5.7 meters = 9.0 meters.

On the other hand, displacement is a vector quantity, which means it has both magnitude and direction. To find the quarterback's displacement, we subtract the southward movement from the northward movement, because these movements are in opposite directions. The displacement is thus 5.7 meters - 3.3 meters = 2.4 meters northward.

Answer:

Matter have two essential types of properties, those are physical properties and chemical properties.

Explanation:

Answer:

d=360 miles

Donna lives 360 miles from the mountains.

Explanation:

Conceptual analysis

We apply the formula to calculate uniform moving distance[

d=v*t   Formula (1)

d: distance in miles

t: time in hours

v: speed in miles/hour

Development of problem

The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:

travel data to the mountains: t₁= 8 hours ,  v=v₁

d= v₁*t₁=8*v₁ Equation (1)

data back home : t₂=4hours ,  v=v₂=v₁+45

d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)

Equation (1)=Equation (2)

8*v₁=4v₁+180

8*v₁-4v₁=180

4v₁=180

v₁=180÷4=45 miles/hour

we replace v₁=45 miles/hour in equation (1)

d=8hour*45miles/hour

d=360 miles

Final answer:

To solve the problem, we used the relationship between speed, distance, and time to set up two equations and solve for the average speed to the mountains and the distance. Donna lives 360 miles away from the mountains.

Explanation:

Let's denote Donna's driving distance to the mountains as D miles. To find this distance, we can use the relationship between speed, distance, and time. When she drove to the mountains, the journey took her 8 hours. For the return trip with no traffic, it took 4 hours and the speed was 45 miles per hour faster.

Let v be the average speed on the way to the mountains. Therefore, the average speed on the way back is v + 45 mph.

Using the formula distance = speed × time, we have two equations:

Since both equations are equal to D, we can set them equal to each other:

v × 8 = (v + 45) × 4

Expanding both sides:
8v = 4v + 180

Subtracting 4v from both sides we get:

4v = 180

Dividing by 4:

v = 45 mph (speed to the mountains)

So, the distance D is:

D = 45 mph × 8 h = 360 miles

Therefore, Donna lives 360 miles away from the mountains.

Answer:

23.63 degree.

Explanation:

Specific heat of water = 4186 joule / kg degree

Specific heat of aluminium  = 900 joule / kg degree

Specific heat of copper = 386 joule / kg degree

Let equilibrium temperature be T.

Heat will be gained by water and aluminium and lost by hot copper.

Heat gained or lost = mst , where m is mass , s is specific heat and t is rise or fall of temperature.

Heat gained by water = .250 X 4186 X ( T - 20 )

= 1046.5 ( T-20)

Heat gained by aluminium = .400 x 900 x ( T - 26 )

= 360 ( T - 26 )

Heat lost by copper = .100 x 386 x ( 100 - T )

38.6 ( 100 - T)

Heat gained = Heat lost

1046.5 ( T-20) + 360 ( T - 26 ) = 38.6 ( 100 - T)

T ( 1046.5 + 360 + 38.6 ) = 3860 + 9360 +20930

1445.1 T = 34150.

T = 23.63 degree.

Change in the entropy of the universe will be zero because no heat is exchanged with the universe. The container is insulated from outside .

Inside the container,  entropy will be increased.  

Final answer:

The student's question about thermal equilibrium and entropy change is a physics problem related to calorimetry in an adiabatic process. It requires calculating the final temperature using heat exchange equations and determining the total entropy change using the formula for entropy change.

Explanation:

The student's question involves the thermal equilibration of three different materials in an insulated container, a common scenario in calorimetry problems within physics. We assume no heat loss to the container or surroundings (adiabatic process), which simplifies the calculations.

Final Temperature Calculation

To find the final temperature, we need to set the heat lost by the warmer substances equal to the heat gained by the cooler substance:

Here, m is mass, c is specific heat capacity, and ΔT is change in temperature.

For water, c = 4186 J/(kg°C), for aluminum, c = 900 J/(kg°C), and for copper, c = 387 J/(kg°C). Solving the equation ∀Q = 0 with the given masses and initial temperatures will yield the final equilibrium temperature.

Entropy Change

The change in entropy is calculated via:

We apply this formula separately for each substance, summing the changes to find the total entropy change of the universe in the experiment.

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

[tex]\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}[/tex]

[tex]2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2[/tex]

[tex]v_2=3.67 m/s[/tex]