Which of the following iS an example of a falsifiable hypothesis that could lead to scientific knowledge?
• A. Loch Ness has a giant reptile living in it.
B. Humans were created.
C. Aliens have crashed on Earth.
D. No aliens have crashed on Earth.

ANSWER : 20 is the answer

Explanation:

If you take 40, then divide it by 2, you get 20. That is how you do the problem.

Can you please mark brainless? It couldn't hurt? STAY SAFE ! :)

Answer:

atmosphere, gravitational pull, magnetic field

Explanation:

the astroids burn up in the atmosphere, gravitational pull alters the path of the astroids, and the magnanetic field deflects them of path.

The angular velocity is 3.75 m/s and the centripetal force is 225 N respectively.

The angular velocity of an object with respect to some extent is a degree of the way rapid that item actions through the point's view, within the feel of the way speedy the angular function of the item modifications. An instance of angular pace is a ceiling fan. One blade will whole a complete round in a certain amount of time T, so its angular speed with respect to the middle of the ceiling fan is twoπ/T.

Calculation:-

A. angular velocity ω = v/r

                                     = 30 /8

                                      = 3.75 m/s

B. Centripetal force = mv²/r

                             = 2×30²/8

                              = 225 N

There are 3 formulations we will use to find the angular velocity. the primary comes instantly from the definition. The angular pace is the rate of alternate of the position attitude of an object with respect to time, so w = theta / t, in which w = angular pace, theta = position attitude, and t = time.

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Answer:

Computer science

Explanation:

You would get everything you need to know about computers if you took computer science. I hope this helps

Answer:

Computer science

Explanation:

1. Diagram and Variables:

                                   Maximum Height

                                          |

                                          |

                                          |

                                          |

                                          |

                                          |

                                          |

                                          |

                                          |

------------------------ Ground ------------------------>

Variables:

Initial velocity (v₀) = 90 m/s

Launch angle (θ) = 45°

Maximum height (H)

Time of travel at maximum height (t_max_height)

Horizontal displacement at maximum height (d_max_height)

Time of travel at maximum range (t_max_range)

Horizontal displacement at maximum range (d_max_range)

2. For when the round reaches maximum height:

a) Time of travel (t_max_height):

At the maximum height, the vertical velocity (v_y) becomes zero. To find the time it takes for the round to reach the maximum height, we can use the equation for vertical motion:

v_y = v₀ * sin(θ) - g * t

0 = v₀ * sin(θ) - g * t_max_height

Solving for t_max_height:

t_max_height = v₀ * sin(θ) / g

Substituting the values:

t_max_height = 90 m/s * sin(45°) / 9.8 m/s²

Calculating the value:

t_max_height ≈ 6.12 s

b) Horizontal displacement (d_max_height):

The horizontal displacement at maximum height can be calculated using the equation:

d_max_height = v₀ * cos(θ) * t_max_height

Substituting the values:

d_max_height = 90 m/s * cos(45°) * 6.12 s

Calculating the value:

d_max_height ≈ 385.94 m

Therefore, at the maximum height, the time of travel is approximately 6.12 seconds, and the horizontal displacement is approximately 385.94 meters.

3. For when the round reaches maximum range:

a) Time of travel (t_max_range):

To find the time it takes for the round to reach the maximum range, we can consider the symmetry of projectile motion. The time of flight (t_flight) is twice the time it takes to reach maximum height:

t_flight = 2 * t_max_height

Substituting the value of t_max_height:

t_max_range = 2 * 6.12 s

Calculating the value:

t_max_range ≈ 12.24 s

b) Horizontal displacement (d_max_range):

The horizontal displacement at maximum range can be calculated using the equation:

d_max_range = v₀ * cos(θ) * t_max_range

Substituting the values:

d_max_range = 90 m/s * cos(45°) * 12.24 s

Calculating the value:

d_max_range ≈ 868.63 m

Therefore, at the maximum range, the time of travel is approximately 12.24 seconds, and the horizontal displacement is approximately 868.63 meters.

When a mortar is fired at an angle of 45 degrees, it will reach its maximum height in 6.49 seconds and its maximum range in 12.98 seconds. The horizontal displacement of the mortar when it reaches its maximum height will be 413.02 meters, and its horizontal displacement when it reaches its maximum range will be 826.53 meters.

1. To draw a diagram of the described scenario, you can start by drawing a coordinate system. The x-axis represents the horizontal direction, and the y-axis represents the vertical direction. Place the origin (0, 0) at the point of launch. Since the mortar is angled 45 degrees from the horizontal, you can draw a line representing the initial direction of the round at a 45-degree angle from the x-axis.

Next, label the variables along the x and y dimensions. For the x-dimension, you can label the variable as "horizontal displacement" or simply "x." For the y-dimension, you can label the variable as "vertical displacement" or "height" and indicate that it is measured in meters.

2. When the round reaches maximum height:

a)

The time of ascent  can be calculated using the following formula:

time = ( initial velocity * sin(angle)) / acceleration due to gravity

In this case, the initial velocity is 90 meters per second, and the angle is 45 degrees. The acceleration due to gravity is typically considered to be approximately 9.8 meters per second squared.

Plugging in the values:

time = (90 * sin(45)) / 9.8  = 6.49s

b) The horizontal displacement at maximum height is :

horizontal displacement = initial velocity * cos (45) * time of ascent

Plugging in the values:

horizontal displacement=90* cos (45) * 6.49s= 413.02m

3. When the round reaches maximum range:

a) The time of travel can be calculated using the following formula:

time = (2 * initial velocity * sin(angle)) / acceleration due to gravity

The initial velocity and angle remain the same.

Plugging in the values:

time = (2 * 90 * sin(45)) / 9.8= 12.98s

b) The horizontal displacement at maximum range can be calculated using the following formula:

horizontal displacement = (initial velocity^2 * sin(2*angle)) / acceleration due to gravity

Plugging in the values:

horizontal displacement = (90^2 * sin(2*45)) / 9.8= 826.53m

Therefore, A mortar will reach its maximum height and distance when shot at a 45-degree angle in 6.49 and 12.98 seconds, respectively. When the mortar achieves its maximum height, its horizontal displacement will be 413.02 meters, and when it reaches its maximum range, it will be 826.53 meters.

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Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

\({ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}\)

Force:

\({ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}\)

Answer:

Explanation:

A) The output force required to lift a 15 kg object would be equal to the weight of the object, which is given by:

Output force = Weight of object = m * g

where m is the mass of the object and g is the acceleration due to gravity. Assuming that g is equal to 9.81 m/s^2, we have:

Output force = 15 kg * 9.81 m/s^2 = 147.15 N

Therefore, the output force required to lift a 15 kg object would be 147.15 N.

B) In this case, the input force is the force that you are pushing down with the lever, which is given as 20 N.

C) The mechanical advantage of the ramp is given by the ratio of the output force to the input force. In this case, the output force is the weight of the object (40 N) and the input force is the force that you are pushing with (10 N). Therefore, the mechanical advantage of the ramp would be:

Mechanical advantage = Output force / Input force = 40 N / 10 N = 4

So, the ramp multiplies your force by a factor of 4.

Note that in all of these calculations, we have assumed that the system is ideal and that there are no losses due to friction or other factors. In practice, these losses will reduce the mechanical advantage of the system and make it more difficult to lift or move objects.

Answer:

v = 0.4 m/s

Explanation:

       \(v_{avg} = \frac{x_{f}-x_{o}}{t_{f} -t_{o}} (1)\)

       \(v_{avg} = \frac{x_{f}}{t_{f} }} (2)\)

       \(v_{avg} =\frac{x_{f}}{t_{f} } = \frac{4.0m}{10s} = 0.4 m/s (3)\)

Given data

*The given equation is

Convert the fraction into decimals as

Answer:

Kr

Explanation:

the element that is in group 18 is noble gasses. the elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

Answer:

Its true

Explanation:

but im answering so the other person can get brainliest :3

To everyone who sees this, have a wonderful day or night!

The rock comes to a complete stop at maximum height, so velocity is zero. Gravity is responsible for the acceleration. As a result, at the top of its path, its velocity is zero and its acceleration is 9.81 m/s2, which is approximately 10 m/s2.

When an object thrown straight upwards gets to the top of its path, Its velocity is zero because it moves opposite to gravity and acceleration is 10 m/s2 , due to the rate of change of velocity of an object with respect to time.

Projectile motion is the movement of an object in which the vertical acceleration is g and the horizontal acceleration is zero. A projectile motion is used to throw a stone into the air. It rises against a deceleration of size g, which slows it down until it reaches the maximum height.

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Explanation:

A push or pull of an objecr

\(\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}\)

A force is described as push or pull on an object.

The answer is C

Please dont get mad at me if this is not right im pretty sure it is
hope this helps!

Answer:

No

Explanation:

S=d/t Speed equals distance divided by time. That's a helpful equation in this case. Leah's wave took 10 seconds, and the distance was 6 meters.

6/10=.6

Betty's wave took 10 seconds, but the distance was 60 meters this time.

60/10=6

We know that 6 and .6 are not the same numbers, so we can see that the waves aren't the same.

<3

Option D. continental shelf

Continental shelf is a part of a continent that is submerged under an area of ​​relatively shallow water known as a shelf sea.

It is given in the question, that the shelf of undersea land reaches 200 meters and it also extends out from the shoreline. This condition is only possible when a structure is submerged, and such submerged region can only be called as continental shelf.

On the other hand guyots is a underwater volcanic mountain and deep-sea trenches are the the long narrow lowerings in the ocean floor with the depth upto 6000 meters. And abyssal plains are the underwater ocean floors which cannot be found at the depth of 200 meters.

Therefore continental shelf is used to described as a shelf of undersea land reaching a depth of about 200 meters (656 feet) and extending out from the shoreline.

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Answer:

Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.

Explanation:

To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.

Given:

m1 = 8 kg

x1 = 3 m

x_cm = unknown (to be found)

m2 = 10 kg

y_cm = 4.5 m

Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.

To solve for x2, we can rearrange the formula as follows:

x2 = (x_cm * (m1 + m2) - m1 * x1) / m2

Substituting the given values:

x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg

Simplifying:

x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg

Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:

4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)

Simplifying:

4.5 m = (24 kg + 10 kg * x2) / 18 kg

Multiplying both sides by 18 kg:

81 kg*m = 24 kg + 10 kg * x2

Subtracting 24 kg from both sides:

10 kg * x2 = 81 kg*m - 24 kg

Dividing both sides by 10 kg:

x2 = (81 kg*m - 24 kg) / 10 kg

Simplifying:

x2 = 8.1 m - 2.4 m

x2 = 5.7 m

(brainlest?) ples:(

Answer:

the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.

Explanation:

To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.

The center of mass of a system is given by the equation:

x_cm = (m1x1 + m2x2) / (m1 + m2),

where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.

Given:

m1 = 8 kg,

x1 = 3 m,

m2 = 10 kg,

y_cm = 4.5 m.

To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:

y_cm = (m1y1 + m2y2) / (m1 + m2),

where y1 and y2 are the positions along the y-axis.

Rearranging the equation and substituting the given values:

4.5 = (83 + 10y2) / (8 + 10).

Simplifying the equation:

4.5 = (24 + 10*y2) / 18.

Multiplying both sides by 18:

81 = 24 + 10*y2.

Rearranging the equation:

10*y2 = 81 - 24,

10*y2 = 57.

Dividing both sides by 10:

y2 = 5.7.

Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.

To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2).

Substituting the given values:

x_cm = (83 + 10x2) / (8 + 10).

Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:

0 = (83 + 10x2) / (8 + 10).

Rearranging the equation:

83 + 10x2 = 0.

24 + 10*x2 = 0.

10*x2 = -24.

Dividing both sides by 10:

x2 = -2.4.

The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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The mass of the proton at the given speed is 2.78 x 10⁻²⁷ kg.

The mass of the proton at the given speed of is calculated by applying the following equation.

Mathematically, the equation relating the mass of the proton and the rest mass is calculated as follows;

m = ( M₀ ) / (√ ( 1 - v² / c² )

where;

The given parameters include;

The mass of the proton at the given speed is calculated as follows;

m = ( M₀ ) / (√ ( 1 - v² / c² )

m = ( M₀ ) / (√ ( 1 - (0.8c)² / c² )

m = M₀ / 0.6

m =  1.67 x 10⁻²⁷ kg / 0.6

m = 2.78 x 10⁻²⁷ kg

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(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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In a tree if the same rock is placed in four different locations, the rock will have the most potential energy.

Potential energy is a form of energy stored that is dependent on the relationship between different system components. When a spring is compressed or stretched, its potential energy increases.

If a steel ball is raised just above ground as opposed to falling to the ground, it has more potential energy. An object has potential energy as a result of its position in relation to other objects. Potential energy is so named because it has the ability to transform into other types of energy, like kinetic energy.

Potential energy is the energy that is held within a material or an object. Energy in an object held vertically is known as gravitational potential energy. Potential energy that can be stretched or compressed is known as elastic potential energy.

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The work done by the Coulomb force will be "\(6.08\times 10^{-21} \ J\)".

Let us define the required work done to move that alpha particle to the one of the mid point of the side length as follows.

→ \(W = \frac{4kqQ}{r_1} -(\frac{2kqQ}{r_2} +\frac{2kqQ}{r_3} )\)

→      \(=2kqQ(\frac{2}{r_1} -\frac{1}{r_2} -\frac{1}{r_3} )\)

→      \(=2(8.99\times 10^9)(-1.6\times 10^{-19})(2)(1.6\times 10^{-19})\)

→      \(= (\frac{2}{\sqrt{(5\times 10^{-9})^2+(5\times 10^{-9})} } -\frac{1}{(5\times 10^{-9})} -\frac{1}{\sqrt{(10\times 10^{-9})^2} +(5\times 10^{-9})^2} )\)

→      \(= 6.08\times 10^{-21} \ J\)

Thus the above answer is appropriate.  

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Answer: 2.07N

Explanation:

∑Fy=4N

∑Fx=0N

The total Tension force on either side must be equal, therefore, Ty+Ty=4N.

Since we are only concerned with the forces in the y direction we will use the sine function.

Ty=Tsinθ

Ty+Ty=2Ty=4N

2Ty=2Tsinθ=4N

Then solve for T,     2Tsinθ=4  ⇒   T= 4/(2sinθ)   ⇒  2.07N

Answer:

A systematic error may result in a high degree of precision.

A systematic error will likely result in poor accuracy

Explanation:

It's what my work gave me when I got it wrong

Answer:

b Day-to-day condition of the atmosphere

Explanation:

Weather is short term, Climate is long term

Answer:

Average atmospheric condition.

Explanation:

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

\(F_e=k\frac{q_1q_2}{r^2}\)            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

\(q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C\)

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

Answer:

input force is force applied to overcome the load and the output force is the force overcome by the effort

Explanation:

effort is equal to input and load is equal to the output

The racer which accelerates at 50.0 m/s2 for 5 seconds then brakes at 60.0 m/s2 to a complete stop has an acceleration of  2 m/s^2, travels a distance of  100m in 5 seconds.

What is acceleration  ?

Acceleration is the measure of a change in velocity. Acceleration frequently, but not always, denotes a shift in speed. The direction of motion is changing, therefore an object travelling at a constant speed along a circular path is still going ahead.

Given:

Acceleration, a=?

Time, t =  5 seconds

Final velocity, v= 60.0 m/s2

Initial velocity, u=  50.0 m/s2

Therefore putting the formula,

v = u + at

=> 60 = 50 +5.a

therefore, a = 2 m/s^2

Therefore distance can be found by the formula,

x = u + 1/2 at^2

Where x is the distance.

=> x = 50 + 2 . 5^2

x = 100m

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