Two of the types of infrared light, IR-C and IR-A, are both components of sunlight. Their wavelengths range from 3000 to 1,000,000 nm for IR-C and from 700 to 1400 nm for IR-A. Compare the energy of microwaves, IR-C, and IR-A.

Answer: Ok, the acceleration decreases means if you have an object with acceleration equal 100 meters for second square, and it starts to decrease slowly to 90, 80... etc, there you still have positive acceleration, so your object's velocity keeps increasing, but more slower than before.

remmember, acceleration means the change in velocity, if you have positive acceleration, your velocity will increase.

the correct answer is D, the air resistance will fight against the gravity, but the object will keep accelerating (with less intensity) ence will go faster and faster.

An object's velocity can increase even as its acceleration decreases, such as in the case of an object falling with air resistance where the rate of speed increase slows down due to the opposing force of air resistance.

It is indeed possible for an object's velocity to increase while its acceleration decreases. This can occur, for example, during a scenario where an object continues to speed up, but the rate at which it is speeding up is decreasing over time. A typical example of this is an object in free fall with air resistance. Initially, as the object begins to fall, it accelerates due to gravity. However, as the object's speed increases, air resistance begins to counteract some of the acceleration due to gravity, causing the acceleration to decrease. Despite the reduction in acceleration, the object's velocity can continue to increase up until it reaches terminal velocity, at which point the acceleration will become zero, but the velocity remains constant and positive.

It is also important to note that acceleration is not always in the direction of motion. When an object is moving and its acceleration is in the direction of motion, it speeds up, while if the acceleration is in the opposite direction of motion, it will slow down or decelerate.

Answer:

Final speed of the racer, v = 16.55m/s

Explanation:

It is given that,

Initial velocity of the racer, u = 12 m/s

Acceleration, [tex]a=0.65\ m/s^2[/tex]

time taken, t = 7 s

We need to find the final velocity of the racer. Let it is given by v. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

[tex]v=12+0.65\times 7[/tex]

v = 16.55 m/s

So, the final speed of the racer is 16.55 m/s. Hence, this is the required solution.

Answer:

A

Explanation:

Forces are vector quantities because they have both magnitude and direction.

A is the right answer i just got it right for A P E X

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.

Know more about normal force,

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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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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:

17.7 days

Explanation:

Since the ship accelerates from the rest, its initial velocity would be equal to 0.

So,

[tex]v_{i}=0[/tex]

Acceleration of the star-ship = a = 1 g = 9.8 m/s²

We need to find how many days will it take the ship to reach 5% of the speed of light. Speed of light is [tex]3 \times 10^{8}[/tex] m/s.

5% of the speed of light = [tex]0.05 \times 3 \times 10^{8}=1.5\times 10^{7}[/tex] m/s

This means, the final velocity of the star-ship will be:

[tex]v_{f}=1.5\times 10^{7}[/tex]

We have the initial velocity, final velocity and the acceleration. We need to find the time(t). First equation of motion relates these quantities as:

[tex]v_{f}=v_{i}+at[/tex]

Using the values in this equation, we get:

[tex]1.5 \times 10^{7}=0+9.8(t)\\\\ t=\frac{1.5\times10^{7}}{9.8}\\\\ t=1,530,612.245[/tex]

Thus, the star-ship will take 1,530,612.245 seconds to reach to 5% the speed of light. Now we need to convert this time to days.

Since, there are 60 seconds in a minute:

1,530,612.245 seconds = [tex]\frac{1,530,612.245}{60}=25510.20[/tex] minutes

Since, there are 60 minutes in an hour:

25,510.20 minutes = [tex]\frac{25,510.20}{60}=425.17[/tex] hours

Since, there are 24 hours in a day:

425.17 hours = [tex]\frac{425.17}{24}=17.7[/tex] days

Thus, it will take approximately 17.7 days (approximately 17 days and 17 hours) to reach to 5% the speed of light

Answer: 23000 frames/s

Explanation:

The rest of the statement of the question is presented below:

The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?

We know the maximum initial speed at which the seeds are dispersed is:

[tex]V_{i}=4,7 m/s[/tex]

In addition, we know the maximum distance at which the seeds move between photographic frames is:

[tex]d_{max}=0.20 mm \frac{1m}{1000 m}=0.0002 m[/tex]

And we need to find the minimum frame rate of the camera with these given conditions. This can be found by finding the time [tex]t[/tex] for each frame and then the frame rate:

Finding the time:

[tex]t=\frac{d_{max}}{V_{i}}[/tex]

[tex]t=\frac{0.0002 m}{4.6 m/s}[/tex]

[tex]t=0.00004347 s/frame[/tex] This  is the time for each frame

Now we need to find the frame rate, which is the frequency at which the photos are taken.

In this sense, frequency [tex]f[/tex] is defined as:

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

[tex]f=\frac{1}{0.00004347 s/frame}[/tex]

Finally:

[tex]f=23000 frames/s[/tex]

Hence, the minimum frame rate is 23000 frames per second.

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.

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.

The net electric force acting on particle 3 due to particle 1 and particle 2 is 4.55 N.

To find the net electric force acting on particle 3 due to particle 1 and particle 2, we can use Coulomb's Law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is: F = k × (|q1 × q3| / r^2)

Where:

Substituting the given values into the formula:

F = (9 × 10^9 Nm^2/C^2) × (|(-8.2 × 10^-9 C) × (8.0 × 10^-9 C)| / (3.3 × 10^-2 m)^2)

Simplifying the equation, we get:

F = 4.55 N

Therefore, the net electric force acting on particle 3 is 4.55 N.

The net electric force acting on particle 3 is approximately [tex]\( 1.62 \times 10^{-3} \, \text{N} \)[/tex] at an angle of [tex]\( 30^\circ \)[/tex] from the horizontal axis.

Given:

[tex]\( q_1 = -8.2 \, \text{nC} \)\\ \( q_2 = -16.4 \, \text{nC} \)\\ \( q_3 = 8.0 \, \text{nC} \)[/tex]

- Side length of the equilateral triangle, [tex]\( a = 3.3 \, \text{cm} = 0.033 \, \text{m} \)[/tex]

Calculate the Magnitude of the Forces

The magnitude of the force between two charges is given by Coulomb's law:

[tex]\[ F = k_e \frac{|q_1 q_2|}{r^2} \][/tex]

where:

- [tex]\( k_e \)[/tex] is Coulomb's constant, [tex]\( k_e = 8.99 \times 10^9 \, \text{N.m}^2/\text{C}^2 \)[/tex]

- r is the distance between the charges, [tex]\( r = 0.033 \, \text{m} \)[/tex]

Force between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex]:

[tex]\[ F_{13} = k_e \frac{|q_1 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-8.2 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{13} = 8.99 \times 10^9 \frac{65.6 \times 10^{-18}}{0.001089} \]\[ F_{13} = 8.99 \times 10^9 \times 6.025 \times 10^{-14} \]\[ F_{13} = 5.41 \times 10^{-4} \, \text{N} \][/tex]

Force between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex]:

[tex]\[ F_{23} = k_e \frac{|q_2 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-16.4 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{23} = 8.99 \times 10^9 \frac{131.2 \times 10^{-18}}{0.001089} \]\[ F_{23} = 8.99 \times 10^9 \times 1.205 \times 10^{-13} \]\[ F_{23} = 1.08 \times 10^{-3} \, \text{N} \][/tex]

Determine the Direction of the Forces

Since the triangle is equilateral, the angles between the forces are 60 degrees. The forces are directed along the lines connecting the charges.

- [tex]\( F_{13} \)[/tex] is directed from [tex]\( q_1 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_1 \)[/tex] is negative, so the force direction is towards [tex]\( q_1 \))[/tex].

- [tex]\( F_{23} \)[/tex] is directed from [tex]\( q_2 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_2 \)[/tex] is negative, so the force direction is towards [tex]\( q_2 \))[/tex].

Calculate the Net Force

We need to resolve the forces into components and sum them.

Resolving [tex]\( F_{13} \)[/tex]:

- Along x-axis: [tex]\( F_{13x} = F_{13} \cos(30^\circ) = 5.41 \times 10^{-4} \cos(30^\circ) \)[/tex]

- Along y-axis: [tex]\( F_{13y} = F_{13} \sin(30^\circ) = 5.41 \times 10^{-4} \sin(30^\circ) \)[/tex]

Resolving [tex]\( F_{23} \)[/tex]:

- Along x-axis: [tex]\( F_{23x} = F_{23} \cos(30^\circ) = 1.08 \times 10^{-3} \cos(30^\circ) \)[/tex]

- Along y-axis: [tex]\( F_{23y} = F_{23} \sin(30^\circ) = 1.08 \times 10^{-3} \sin(30^\circ) \)[/tex]

Since both forces have the same y-component direction:

[tex]\[ F_{13x} = 5.41 \times 10^{-4} \times \frac{\sqrt{3}}{2} \approx 4.68 \times 10^{-4} \, \text{N} \]\[ F_{13y} = 5.41 \times 10^{-4} \times \frac{1}{2} \approx 2.70 \times 10^{-4} \, \text{N} \]\[ F_{23x} = 1.08 \times 10^{-3} \times \frac{\sqrt{3}}{2} \approx 9.35 \times 10^{-4} \, \text{N} \]\[ F_{23y} = 1.08 \times 10^{-3} \times \frac{1}{2} \approx 5.40 \times 10^{-4} \, \text{N} \][/tex]

The net force components are:

[tex]\[ F_{net_x} = F_{13x} + F_{23x} = 4.68 \times 10^{-4} + 9.35 \times 10^{-4} \approx 1.40 \times 10^{-3} \, \text{N} \]\[ F_{net_y} = F_{13y} + F_{23y} = 2.70 \times 10^{-4} + 5.40 \times 10^{-4} \approx 8.10 \times 10^{-4} \, \text{N} \][/tex]

Calculate the Magnitude of the Net Force

[tex]\[ F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2} = \sqrt{(1.40 \times 10^{-3})^2 + (8.10 \times 10^{-4})^2} \]\[ F_{net} = \sqrt{1.96 \times 10^{-6} + 6.56 \times 10^{-7}} \]\[ F_{net} = \sqrt{2.62 \times 10^{-6}} \]\[ F_{net} \approx 1.62 \times 10^{-3} \, \text{N} \][/tex]

Determine the Direction of the Net Force

The direction of the net force can be found using the angle:

[tex]\[ \theta = \tan^{-1} \left( \frac{F_{net_y}}{F_{net_x}} \right) = \tan^{-1} \left( \frac{8.10 \times 10^{-4}}{1.40 \times 10^{-3}} \right) \]\[ \theta = \tan^{-1} (0.579) \]\[ \theta \approx 30^\circ \][/tex]

Yes. According to the coach's mathematical criteria, Tommy had a great season.

Sadly, Tommy doesn't even know it. His tone-deaf coach decided to describe success in terms that are absurd for 7-yr-olds, as well as for most of their parents.

Answer:

Q=22.6L/s

Explanation:

First you must consider the continuity equation at points 1 and 2, which indicates that both flows are of equal value, in this way you get an equation between the two flow rates.

Then you raise the Bernoulli equation taking into account that the height is the same, which makes the term h1-h2 zero.

Using the equations above to calculate one of the speeds.

Finally you find the flow by multiplying the speed by the area.

I attached procedure

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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Answer:

New force, F' = F

Explanation:

Given that, two small balls, A and B, attract each other gravitationally with a force of magnitude F. It is given by :

[tex]F=G\dfrac{m_Am_B}{r^2}[/tex]

If we now double both masses and the separation of the balls, the new force is given by :

[tex]F'=G\dfrac{2m_A\times 2m_B}{(2r)^2}[/tex]

F' = F

So, the new force remains the same as previous one. Hence, the correct option is (d) "F"

Answer:

After studying the law of gravitational attraction, students constructed a model to illustrate the relationship between gravitational attraction (F) and distance. If the distance between two objects of equal mass is increased by 2, then the gravitational attraction (F) is 1/4F or F/4. How would this model, situation A, change if the mass of the spheres is doubled?

A)  A

B)  B

C)  C

D)  D

If you came here from usa test prep it is:

Actually A

But for the question given right now is D.

Explanation:

Answer:

2.76m/s

Explanation:

The position x for a given time t and constant acceleration a is:

(1) [tex]x=\frac{1}{2}at^2[/tex]

The velocity v:

(2) [tex]v=at[/tex]

Solving equation 2 for time t:

(3) [tex]t=\frac{v}{a}[/tex]

Combining equations 1 and 3:

(4) [tex]x=\frac{v^2}{2a}[/tex]

Solving equation 4 for velocity v:

(5) [tex]v=\sqrt{2ax}[/tex]

For a = 0.465m/s² and x = 8.2m:

v = 2.76m/s

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

Answer: x = 0.237

y = -0.0789

Answer:

[tex]v_o = 2761 m/s[/tex]

Explanation:

Since there is no external Force or Torque on the system of bullet and rod

So we can say that total angular momentum of the system will remain conserved about its center

So we will have

[tex]mv_o\frac{L}{2}sin\theta = (I_{rod} + I_{bullet})\omega[/tex]

here we know that

[tex]I_{rod} = \frac{mL^2}{12}[/tex]

[tex]I_{rod} = \frac{4.10\times 0.70^2}{12} [/tex]

[tex]I_{rod} = 0.167 kg m^2[/tex]

[tex]I_{bullet} = mr^2[/tex]

[tex]I_{bullet} = (0.003)(0.35^2)[/tex]

[tex]I_{bullet} = 3.675 \times 10^{-4} kg m^2[/tex]

[tex]\omega = 15 rad/s[/tex]

[tex]\theta = 60 [/tex]

now we have

[tex]0.003(v_o)(0.35)sin60 = (0.167 + 3.675 \times 10^{-4})15[/tex]

[tex]v_o = 2761 m/s[/tex]

This physics question is about calculating the speed of a bullet before it strikes a rotating rod, based on the conservation of angular momentum. After setting up the equations for initial and final angular momentum, by equating these two, we can find the speed of the bullet just before it hits the rod.

This is a physics problem regarding the conservation of angular momentum during a collision. We are given a scenario where a bullet impacting and lodging into a rotating rod which is initially at rest. According to the conservation of angular momentum, the initial angular momentum prior to the collision should equal the angular momentum after the collision, if no external torque is acting on the system.

The initial angular momentum (just before the collision) is the product of the bullet's mass, speed, and distance from the axis of rotation (which is half of the rod's length), and the cos(θ), where θ is the angle the bullet's path makes with the rod. In this case, angular momentum is mbrvb cos(θ)r

The final angular momentum (just after the collision) is the moment of inertia of the system (bullet plus rod) times the ensuing angular velocity, which is (mb(r^2) + (1/12)M(r^2))ω

By setting the initial and the final angular momentum equations equal to each other and arranging for vb, we will find the speed of the bullet just before impact.

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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:

[tex]E_{th} = 19680 J[/tex]

Explanation:

GIVEN DATA:

Total mass ( rider  + his tube) = 80 kg

tension  Force = 360 N

height of slope  =  30 m

length of slope = 120 m

we know that thermal energy is given as

E_{th}  = W- Ug

W= F*d = (360N*120m)= 43200 J        

Ug= m*g*h = (80kg*9.8m/s2*30m) = 23520 J

[tex]E_{th} = 43200J - 23520 J[/tex]

[tex]E_{th} = 19680 J[/tex]

The amount of thermal energy that is created in the slope and the tube during the ascent is 19680 Joules.

Given the following data:

To find the amount of thermal energy that is created in the slope and the tube during the ascent:

By applying the Law of Conservation of Energy:

[tex]Work\;done = K.E + P.E + T.E[/tex]

Where:

Since the rider and his tube are pulled at a constant speed, K.E is equal to zero (0).

Therefore, the formula now becomes:

[tex]Work\;done = 0 + P.E + T.E\\\\T.E = Work\;done-P.E[/tex]

For the work done:

[tex]Work\;done = Tensional\;force \times displacement\\\\Work\;done = 360 \times 120[/tex]

Work done = 43,200 Joules.

For P.E:

[tex]P.E = mgh\\\\P.E = 80\times9.8\times30[/tex]

P.E = 23,520 Joules.

Now, we can find the amount of thermal energy:

[tex]T.E = Work\;done-P.E\\\\T.E =43200-23520[/tex]

T.E = 19680 Joules.

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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.

Answer:

The building is 20.68m high.

Explanation:

In order to solve this problem we must first do a drawing of what the situation looks like. (Look at the attached picture). Then, we determine what was the velocity of the ball when it reached the top of the window (let's call this [tex]V_{0A}[/tex] and when it reached the bottom of the window (let's call this [tex]V_{0B}[/tex]. We can start by finding the velocity the ball has at the top of the window [tex]V_{oA}[/tex] when it's falling.

In order to do so, we can use the following equation:

[tex]\updelta y=V_{0A}t+\frac{1}{2} at^{2}[/tex]

We can now solve this equation for [tex]V_{0A}[/tex] so we get:

[tex]v_{0A}=\frac{\updelta y-\frac{1}{2} at^{2}}{t}[/tex]

or when simplified we get:

[tex]v_{0A}=\frac{2\updelta y-at^{2}}{2t}[/tex]

now we can substitute values in the formula so we get:

[tex]v_{0A}=\frac{2(-1.20m)-(-9.8m/s^{2})(0.14s)^{2}}{2(0.14)}[/tex]

and solve:

[tex]v_{0A}=-7.886m/s[/tex]

Once we got the first velocity, we can proceed and find the second velocity.

We can find that by using the following formula:

[tex]a=\frac{V_{f}^{2}-V_{0}^{2}}{2y}[/tex]

So now we can solve for [tex]V_{f}[/tex]

When solving for [tex]V_{f}[/tex], we get the following formula:

[tex]V_f=\sqrt{2ya+V_{0}^{2}}[/tex]

So we can now substitute some values.

[tex]V_{0B}=\sqrt{2(-1.20m)(-9.8m/s\{2})+(-7.885m/s)^{2}}[/tex]

When solving this equation we get an answer of:

[tex]V_{0B}=-9.257m/s[/tex]

Once we got those velocities, we can use them to find the distance from the roof to the highest part of the window and the distance between the

lowest part of the window and the floor.

Let's start with the top portion of the window:

[tex]y=\frac{V_{f}^{2}-V_{0}^{2}}{2A}[/tex]

in this case, the initial velocity is 0 because we are dropping the ball from the roof. So we can now substitute values so we get:

[tex]y=\frac{(-7.885m/s)^{2}-(0)^{2}}{2(-9.8m/s^{2}}[/tex]

when solving this for y we get:

y=-3.17m

Now, we can find the height between the lowest part of the window and the floor, so we get:

[tex]\updelta y=V_{0B}t+\frac{1}{2} at^{2}[/tex]

In this case the time is 1.11s because it's half of the total time the ball is beneath the window when it's falling. So when substituting values we get:

[tex]\updelta y=(-9.257m/s^{2})(1.11s)+\frac{1}{2} (-9.8m/s^{2})(1.11s)^{2}[/tex]

When solving that expression we get that:

[tex]\updelta y=-16.31m[/tex]

so now we have enough information to solve the problem, notice that the heights appear as negative. This is because the bal is falling at this time. Since we only care about the magnitud, we can make them positive.

So in order to find the height of the building, we must ad the three lengths we just found, so we get:

h=3.17m+1.20m+16.31m=20.68m

So the building has a height of 20.68m

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:

4.933m/s

Explanation:

the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N

of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.

then, the force that is moving the wagon is 125N-108N=17N

F=m*a then 17N=35.1kg*a

a=0.4843 m/s2

we have two equations

[tex]v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}[/tex]

then

[tex]x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}[/tex]

[tex]t=\sqrt{\frac{2x}{3a} }[/tex]

t=10.188s

[tex]v=a.t[/tex]

v=4.933 m/s

Answer:

True

Explanation:

Distance is defined as the length of the actual path traveled by the body.

Displacement is defined as the minimum distance between the two points.

the magnitude of displacement is always less than or equal to the distance traveled by the body.

As a deer runs from A to b , so it means the distance traveled by the deer is either equal to the magnitude of displacement or always greater than the magnitude of displacement of the deer.

Displacement can never be greater than the distance.

Thus, the option is true.

Answer:

lower vapor pressure and freezing point and a higher boiling point

Explanation:

All the properties listed above are colligative properties, they depend on the amount of solute present in a solution.

It is known that a solution has a lower vapour pressure than the corresponding pure solvent due to the presence of a solute. Also the freezing point of a solution is decreased while its boiling point is increased when compared with those of the pure solvent.

A salt water solution, compared to pure water, has a lower vapor pressure and freezing point, but a higher boiling point due to the impact of the salt, a non-volatile solute.

Compared to pure water, a salt water solution has lower vapor pressure, lower freezing point, and a higher boiling point. The presence of salt, a non-volatile solute, reduces the vapor pressure of the solution, leading to a decreased freezing point and an increased boiling point. This is due to a phenomenon known as colligative properties, which states that adding a solute to a solvent affects the physical properties of the solvent, regardless of the nature of the solute.

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Answer:

t=0.47s

Explanation:

the ball has uniformly accelerated movement due to gravity

Vo=initial speed=4.6m/s

g=gravity=-9.8m/s^2

Vf=final speed=0, the player must wait for the ball to stop. so the final speed will be 0

we can use the following ecuation

T=(Vf-Vo)/g

T=(0-4.6)/-9.8m/s^2

T=0.47s

Answer:

[tex]v = 33.66 m/s[/tex]

Explanation:

Let hockey puck is moving at constant speed v

so here we have

[tex]d = vt[/tex]

so time taken by the puck to strike the wall is given as

[tex]t = \frac{58.2}{v}[/tex]

now time taken by sound to come back at the position of shooter is given as

[tex]t_2 = \frac{58.2}{340}[/tex]

[tex]t_2 = 0.17s[/tex]

so we know that total time is 1.9 s

[tex]1.9 = t + t_2[/tex]

[tex]1.9 = t + 0.17[/tex]

[tex]1.9 - 0.17 = t[/tex]

[tex]t = 1.73 s[/tex]

now we have

[tex]1.73 = \frac{58.2}{v}[/tex]

[tex]v = 33.66 m/s[/tex]

Answer:

0.146 m/s

Explanation:

We can see it in the pic.

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.