What happens when light from the sun passes through any type of matter

The amount of time the ball stay in the air is (t)=3.04 Seconds

The best example that would help us understand and know what time are the clock and the calendar. This clock gives us the exact hour, minutes and seconds. The calendar tells us the exact day, month and year.

To calculate the amount of time the ball stay in the air, we are using the formula,

T=2Vsinθ/g

Here we are given,

V= The velocity of the football.=26 m/s.

θ= The angle that the football makes in the air= 35°

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

We have to calculate, the amount of time the ball stay in the air = T

Now, we put the known values in the above equation,

T=2Vsinθ/g

Or, T= 2*26*sin (35°)/9.80

Or, T= 3.04 Seconds.

Thus, from the above calculation we can conclude that the amount of time the ball stay in the air is (t)=3.04 Seconds

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Answer:

potential energy

Explanation:

The bent ruler possesses elastic potential energy.  

• The bending of the ruler depends upon the material of which it is made up of.  

• The ruler after bending will gain elastic potential energy.  

• The energy stored as an outcome of applying a force to deform an elastic object is known as the elastic potential energy.  

• This energy is stored unless and until the removal of force takes place and the ruler moves back to its original shape.  

• When the energy is released or when the bent ruler is released from the left hand, then the potential energy stored in it gets convert to kinetic energy and the ruler will recoil or vibrate briskly.  

Thus, it is the elastic potential energy gained by the ruler when one bent it with the left hand.

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The amount of work required to slide the 3 Kg box 6.90 m along (up) the frictionless incline plane is 183.85 J

Work is defined as the product of force and distance moved in the direction of the force. mathematically it is expressed as:

Work done (Wd) = force (F) × distance (d)

Wd = fd

Where height is involved, the workdone is given as as:

Work done (Wd) = mass (m) × acceleration due to graity (g) × height (h)

Wd = mgh

Considering angle projection,

Wd = mghCosθ

Thus, we can detertmine the work done as follow:

Wd = mghCosθ

Wd = 3 × 9.8 × 6.9 × Cos 25

Wd = 183.85 J

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The distance that automobile travel in 1.5 seconds is (S)= 34.275 m.

Distance is a scaler quantity that refers to "how much ground an object has covered" during its motion of time.

To calculate the distance that automobile travel in 1.5 seconds, we are using the formula,

S= ut+(1/2)at²

Here we are given,

u= The initial velocity of the automobile = 20 m/s

t = The amount of time that the automobile travel = 1.5 Seconds.

a=The amount of acceleration of the automobile = 3.8 m/s².

We have to calculate the distance that automobile travel in 1.5 seconds = S

Now, we put the values in the above equation, we get

S= ut+(1/2)at²

Or, S= 20*1.5+(1/2)*3.8*(1.5)²

Or, S= 30+4.275

Or, S= 34.275m

Thus we can conclude that, the distance that automobile travel in 1.5 seconds is (S)= 34.275 m.

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Answer:

3400 m

Explanation:

Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;

 speed = distance/time, hence, distance = speed x time.

For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as;

distance = 340 m/s x 10 s = 3400 m

Answer:

Image B

Explanation:

although I'm not exactly sure, i've recently gotten this question as well. but model B demonstrates the force- distance trade off because you can see how in that image them distance is increased in the force is decreased with the object being shorter. hopefully this helps in some way

Given the speed of the driver and the time elapsed, distance traveled during his inattentive period is 0.037 kilometers.

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

First, convert 1.7 seconds to hours.

Time = 1.7s = (1.7 / (60×60)hr = 1.7/3600 hrs

Now, find the distance traveled during the inattentive period.

80 km/h = Distance ÷ 1.7/3600 hrs

Distance = 80 km/h × 1.7/3600 hrs

Distance = 136/3600 km

Distance = 0.037 km.

Given the speed of the driver and the time elapsed, distance traveled during his inattentive period is 0.037 kilometers.

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Answer:

The force of friction acts in the direction opposite to the direction of motion. If friction would have been applied to the skier it would have resulted in a lower velocity and less kinetic energy.

Explanation:

Answer:

Explanation:

.... ...

Answer:

8

Explanation:

The energy levels is given as

E(n) = n² * h² / ( 8 * m * L²), where

n = 1, 2, 3, 4,... etc

At ground state energy (n = 1), therefore is E(g) = h² / (8 * m * L²).

We can then say that

E(n) = n² * Eg

Therefore, to have E = 64 Eg, we must have n² = 64. And for n² to be equal to 64, we find the square root of n

n = √64

n = 8

Essentially, the needed quantum number is 8

Golf carts have a high velocity.

Without any improvements, the conventional golf cart's peak velocity is between 12 and 14 miles per hour. The highest velocity that you can get on a golf cart without any upgrades or modifications is around 14 mph.

We can determine how quickly or slowly an item is traveling using the phrases velocity and speed. It occurs to us frequently that we must determine which of two or more moving things is traveling quicker. If both vehicles are traveling in the same direction along the same route, it is simple to determine which is quicker. It is challenging to identify who is moving quickest when their direction of motion is opposing. The idea of velocity is useful in these situations.

When gravity first exerts force on an item, its initial velocity defines how quickly the thing moves. The final velocity, on the other hand, is a vector number that gauges a moving body's speed and direction after it has attained its maximum acceleration.

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Answer:

B. 2 Hz

Explanation:

Given parameters:

Period of the wave  = 0.5Hz

Unknown:

Frequency of the wave  = ?

Solution:

Frequency is the number of waves that passes through a point.

Period is the time taken for a number of waves to pass through a point.

  Period is the inverse of frequency

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

 t is the period

  Frequency  = [tex]\frac{1}{0.5}[/tex]  = 2Hz

Answer:2

Explanation:

Answer:

1) 50°

Explanation:

We need to find the angle of incidence first before finding the angle of reflection.

Angle of incidence = 90° - 40°

                                = 50°

Since the angle of incidence is the same as the angle of reflection, the angle of reflection here would be 50°.

Answer:

0 j

Explanation:

The work done by the employee on the box at the given zero displacement is 0 J.

The given parameters;

The work done by the employee on the box is calculated as follows;

W = Fd

where;

Since the employee sits on the box without moving it, the distance moved by the box = 0

W = 500 x 0

W = 0 J

Thus, the work done by the employee on the box is 0 J.

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Answer:

x =  0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

[tex]Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2[/tex]

Now, we will calculate the force on the piston due to atmospheric pressure:

[tex]Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N[/tex]

Now, for the compression of the spring we will use Hooke's Law as follows:

[tex]F = kx\\[/tex]

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

x =  0.0734 m = 7.34 cm

The direction of the current from the positive to the negative battery terminals and the fact that the bulbs are connected in parallel to each other suggests that this circuit is a Closed Parallel Circuit. The correct option is D.

An open circuit and a closed circuit are two types of electrical circuits that describe the flow of electrical current through a circuit.

An open circuit is a circuit in which there is a break in the path of the current, meaning that the current cannot flow through the circuit. In an open circuit, the switch is in the "off" position or there is a broken wire, and so no electrical current can flow through the circuit. This means that there is no voltage or electrical energy being transferred from the source to the load.

On the other hand, a closed circuit is a circuit in which there is a complete path for the current to flow through, meaning that the current can flow through the circuit. In a closed circuit, the switch is in the "on" position, and there is a continuous path for the current to flow from the source to the load and back to the source. This means that there is voltage and electrical energy being transferred from the source to the load.

Here in this question,

The direction of the current from the positive to the negative battery and the fact that the bulbs are connected in parallel to each other suggests that this circuit is a Closed Parallel Circuit.

In a closed parallel circuit, the components are connected in parallel to each other, meaning that they are connected to the same two points in the circuit and the current has multiple paths to flow through. The voltage across each component is the same, and the total current flowing through the circuit is divided among the components according to their resistance.

In contrast, an open series circuit is a circuit in which the components are connected in series, meaning that they are connected end to end in a single path, with no other branch points for the current to follow. If one component fails, the circuit becomes open and the current stops flowing.

Therefore, based on the information provided, the circuit described in the question is a Closed Parallel Circuit.

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Answer:

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

Answer:

[tex]Fr_k=152.88\ N[/tex]

Explanation:

Net Force

The net force is defined as the vector sum of all the forces acting on a body at a certain moment.

We should recall some basic concepts and equations to solve the problem.

If no external forces are applied in the vertical direction, the weight of the object and the normal force have the same magnitude and point to opposite directions.

The friction force is defined as:

[tex]Fr_k=\mu_k N[/tex]

[tex]Fr_s=\mu_s N[/tex]

Where the subindices k and s are referred as to the kinetic and static friction forces respectively.

The condition for the object to move is that the applied force is greater than the friction force.

The crate of oranges has a mass of 30 Kg, thus its weight is:

W = m.g = 30 * 9.8 = 294 N

The normal force is:

N = W = 294 N

The kinetic friction is calculated as:

[tex]Fr_k=0.52* 294[/tex]

[tex]\mathbf{Fr_k=152.88\ N}[/tex]

density of object is 6.7g/cm³  and volume is 30cm³.

First, air has a low density, so the mass in air is very close to the true mass (mass in a vacuum). Let's say the true mass is 200g.

Then we can use the following formula to find the volume:

AM=TM-V•D where AM is the apparent mass (in water) and TM is the true mass (mass in vacuum which approximates mass in air). , V is the volume of water displaced by the object (which is also the volume of the object) and D is the density of the water, i.e.

170g=200g-V*(1,000g/cm3)

V=30cm³

Therefore The density of is approximately 200g÷30cm³=6.7g/cm³

If you want an exact answer, you can find the density of air under laboratory conditions. For example,

1.20g/L or 0.00120g/cm³. Then, using the above equations, we can set up a system of two equations with two unknowns.

200g=TM-V*(0.00120 g/cm³)

170g=TM-V*(1.000 g/cm³)

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Answer:

I got this question on Ap3x. The answer is Car C...I got it correct

Explanation:

This is because Car C is at the lowest point with the lowest amount of potential energy. Potential energy is stored at it's highest when it has the "potential" to fall, move, or etc. Car C seems to have gone farther down from the high point of the slope, meaning that most of the potential energy transformed into kinetic energy. All in all, Car C has the least potential energy. (Please give Brainliest, or not...your choice but this is the first question that I answered)

Answer:

6sec and 8 postion

Explanation:

Answer:

Approximately [tex]1.69 \times 10^{-23}\; {\rm N}[/tex].

Explanation:

Look up the value of Coulomb's Constant: [tex]k \approx 8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}}[/tex].

Consider point charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. If the distance between these charges is [tex]r[/tex], the magnitude of the electrostatic force between them would be [tex](k\, q_{1}\, q_{2}) / (r^{2})[/tex].

In this question, the two [tex](+q)[/tex] charges are [tex]5\; {\rm cm}[/tex] and [tex]3\; {\rm cm}[/tex] away from the center [tex](-2\, q)[/tex] charge, respectively. Convert units to standard unit of distance (meters, [tex]{\rm m}[/tex]) and charge (coulombs, [tex]{\rm C}[/tex]):

[tex]q = 1.15 \; {\rm nC} = 1.15 \times 10^{-9}\; {\rm C}[/tex].

[tex]\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}[/tex].

[tex]\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}[/tex].

The magnitude of the electrostatic forces on the [tex](-2\, q)[/tex] charge would be:

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.05\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 9.509\times 10^{-24}\; {\rm N}\end{aligned}[/tex].

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.03\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 2.641\times 10^{-23}\; {\rm N}\end{aligned}[/tex].

Since the charges are of opposite sign, the [tex](-2\, q)[/tex] charge would attract both of the [tex](+q)[/tex] charges. In particular, the (approximately) [tex]9.509\times 10^{-24}\; {\rm N}[/tex] force would point to the left. The (approximately) [tex]2.641 \times 10^{-23}\; {\rm N}[/tex] force would point to the right.

As a result, the net force on the [tex](-2\, q)[/tex] charge would point to the right. The magnitude of the net force on this charge would be approximately [tex]2.641 \times 10^{-23}\; {\rm N} - 9.509\times 10^{-24}\; {\rm N} \approx 1.69 \times 10^{-23}\; {\rm N}[/tex].

Answer:

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Explanation:

In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:

[tex]N = \frac{12\,oz}{12\,\frac{oz}{dr} }+\frac{12\,oz}{5\,\frac{oz}{dr} }+\frac{3\,oz}{1.5\,\frac{oz}{dr} }[/tex]

[tex]N = 1\,dr+2.4\,dr+2\,dr[/tex]

[tex]N = 5.4\,dr[/tex]

[tex]N = 6\,dr[/tex]

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Answer:

C

Explanation:

If the arrows represent light rays, then Rachel sees a candle flame when the light released by the flame is received by her eyes.

Diagram C shows how Rachel can see a candle flame.

This is the visual representation of a substance and the organ responsible for this is the eye.

Images are formed through the eyes receiving light sensations and and relaying it to the brain for processing. The line is the light sensation which shows it being received by the eye therefore making option C the most appropriate option.

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Answer:

A Living things can only come from living things.

B Cells come from pre-existing cells.

Explanation:

sorry if wrong e d g e 2022

Answer: D,E

Explanation:

I JUST TOOK IT

Answer:

The weight of the brick is 300 lb.

Explanation:

Given that,

The work done in lifting a load of bricks = 15000 ft lb

It is lifted to a height of 50 ft

We need to find the weight of the brick. The work done by an object is given by :

[tex]W=F\times d[/tex]

Here, F = W (weight of the bricks)

[tex]F=\dfrac{W}{d}\\\\F=\dfrac{15000\ ft-lb}{50\ ft}\\\\F=300\ lb[/tex]

So, the weight of the brick is 300 lb.

The weight of the brick will be "300 lb".

Given:

Work done,

Height,

As we know the formula,

→ [tex]W = F\times d[/tex]

or,

→ [tex]F = \frac{W}{d}[/tex]

By substituting the values, we get

→     [tex]= \frac{15000}{50}[/tex]

→     [tex]= 300 \ lb[/tex]

Thus the response above is right.  

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Explanation:

Objects at rest and in motion respond to the presence of an external unbalanced force by simple changing their magnitude of motion or position.

We have this knowledge from Newtons first law of motion "a body will remain in a state of rest or continue with uniform motion unless if it is acted upon by an external force".

Answer:

The acceleration acting on the train during this

time of travel is 0.85m/s²

HOW TO CALCULATE ACCELERATION:

• The acceleration of a moving body can be calculated by using the formula below:

a = (v - u)/t

Where;

1. a = acceleration (m/s²)

2. v = final velocity (m/s) 3. u = initial velocity (m/s)

4. t = time (s)

According to this question, a = ?, v = 48m/s, u = 14m/s, t = 40s.

= (48 - 14)/40 a=

• a = 34/40

• a = 0.85m/s²

Therefore, the acceleration acting on the train during this time of travel is 0.85m/s².

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