Is a scientific law more accurate than a theory?

The work done by the teacher can be calculated using the equation:

work = force x distance x cos(θ)

Here, the force is 10 N, the distance is 20 feet (converted to meters) and the angle between the direction of force and distance is 0 degrees, so cos(0) = 1.

So, work = 10 N x 20 feet x 0.3048 m/foot x 1 = 60.96 J

The teacher did 60.96 joules of work.

Answer:

Warm colours are those with a yellow-ish or black undertone. Imagine how everything looks through your tinted sunglasses or on a beautiful day in autumn when the leaves are falling. The hints of yellow give these colours a sunny warmth, while those with more black mixed in appear darker and can seem more muted.

Answer:

Work = Force * Distance

Here,

force = 124 N

distance = 13m

Thus, Work = 124 * 13 = 1612 joules(J)

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

The resultant velocity of the golf ball is 18.03 m/s

What is Resultant velocity?

An object's total vector velocity is the sum of its component vector velocities. The scalar product of an object's mass and acceleration vector equals the sum of the vector forces acting on it. If indeed the two objects stick together again after impact, divide the total momentum even by sum of the masses. You will then be given the two objects' resultant velocity as a result. In the aforementioned example, 50 would be divided by the total mass of the masses, which really is 10, yielding a speed of 5 metres per second.

Horizontal velocity= 10 m/s

Vertical velocity= 15 m/s

Resultant velocity= \(\sqrt{100+225}\) = \(\sqrt{325}\) = 18.03 m/s

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tTe spring constant of the spring, attached with when a mass of 0.20 kg  is 0.059 N/m.

The spring constant, k, is a measure of the stiffness of the spring.

To calculate the spring constant of the spring, we use the formula below.

Formula:

Where:

From the question,

Given:

Susbtitute these values into equation 1

Hence, the spring constant of the spring is 0.059 N/m.

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The boxes acceleration up the ramp is approximately 0.109 m/s².

The forces acting on the box can be resolved into two components: one parallel to the incline (F_parallel) and one perpendicular to the incline (F_perpendicular).

Given:

Mass of the box (m) = 40.7 kg

Coefficient of kinetic friction (μ) = 0.17

The force pulling the box (F_p) = 173 N

Incline angle (θ₁) = 14.5°

Force angle (θ₂) = 25.7°

First, we need to calculate the components of the force pulling the box:

F_parallel = F_p * sin(θ₂)

F_perpendicular = F_p * cos(θ₂)

Next, let's calculate the force of friction:

F_friction = μ * (mass of the box) * g, where g is the acceleration due to gravity (approximately 9.8 m/s²)

The force component parallel to the incline is opposed by the force of friction, so:

Net force parallel to the incline (F_net_parallel) = F_parallel - F_friction

Now, we can calculate the acceleration using Newton's second law:

F_net_parallel = (mass of the box) * acceleration

Rearranging the equation, we get:

acceleration = F_net_parallel / (mass of the box)

Now we can substitute the values into the equations and calculate the acceleration

F_parallel = 173 N * sin(25.7°) ≈ 73.88 N

F_perpendicular = 173 N * cos(25.7°) ≈ 154.37 N

F_friction = 0.17 * (40.7 kg) * 9.8 m/s² ≈ 69.44 N

F_net_parallel = F_parallel - F_friction ≈ 73.88 N - 69.44 N ≈ 4.44 N

acceleration = (4.44 N) / (40.7 kg) ≈ 0.109 m/s²

Therefore, the box's acceleration up the ramp is approximately 0.109 m/s².

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Element B has 12 electrons. Thus, its atomic number is 12 and it is magnesium (Mg). Its group members with same number of valence electrons are Ca and Sr. C with 4 energy levels is Sr and A is Ca.

Electrons in an atom is filled from the lowest energy level to the highest energy level. The electrons filled in the lower energy levels in the inner shells are called inner electrons pr core electrons and the electrons filled in the outer orbital or valence shell are called the valence electrons.

Given that element B has 12 electrons. The element with 12 electrons is magnesium (Mg). Elements of the same group have same number of valence electrons.  The group members of Mg are Ca, Sr, Rb etc.

Here the element B is lightest among the three. Hence, the element C is just after B with same number of energy levels and it is calcium Ca. The element C with 4 energy level is strontium (Sr).

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a.  The acceleration of the 4.00 kg crate is 3.75 m/s^2.

b.  A free-body diagram for the 4.00 kg crate is shown below:

4.00 kg

T

c. The direction of the force is to the right, since the crate is moving to the right. The magnitude of the force is 15 N.

Acceleration can be described as  the rate of change of the velocity of an object with respect to time.

We use  Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration

(F_ = m * a).

The total force on each crate is equal to the force applied on it. If the force applied on the 6.00 kg crate produces an acceleration of 2.5 m/s^2, the  total force on the 6.00 kg crate is 6.00 kg * 2.5 m/s^2 = 15 N.

Therefore, the total  force on the 4.00 kg crate is 4.00 kg * a = 15 N,

a = 15 N / 4.00 kg

a = 3.75 m/s^2.

The tension T in the rope that connects the two crates can be found using Newton's second law of motion.

b.  (F = m * a), so we can put out the  the equation as :

T - 4.00 kg * 3.75 m/s^2 = 0

T = 4.00 kg * 3.75 m/s^2 = 15 N

c.

F - 6.00 kg * 2.5 m/s^2 = 0

F = 6.00 kg * 2.5 m/s^2 = 15 N

The direction of the force is to the right, since the crate is moving to the right. The magnitude of the force is 15 N.

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Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.

Veremos que la longitud del nuevo péndulo debe ser 6.25m

Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.

La frecuencia de un péndulo está dada por:

\(f = \frac{1}{2*\pi} *\frac{g}{l}\)

Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:

\(5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2\)

Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:

\(4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2} = 6.25m\)

La longitud del nuevo péndulo deve ser 6.25m

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There is more to the Earth than what we can see on the surface. In fact, if you were able to hold the Earth in your hand and slice it in half, you'd see that it has multiple layers. But of course, the interior of our world continues to hold some mysteries for us. Even as we intrepidly explore other worlds and deploy satellites into orbit, the inner recesses of our planet remains off limit from us.

However, advances in seismology have allowed us to learn a great deal about the Earth and the many layers that make it up. Each layer has its own properties, composition, and characteristics that affects many of the key processes of our planet. They are, in order from the exterior to the interior – the crust, the mantle, the outer core, and the inner core. Let's take a look at them and see what they have going on.

Like all terrestrial planets, the Earth's interior is differentiated. This means that its internal structure consists of layers, arranged like the skin of an onion. Peel back one, and you find another, distinguished from the last by its chemical and geological properties, as well as vast differences in temperature and pressure.

Explanation:

The five examples of pseudoscience found on the Internet include the following:

Pseudosciences is defined as the science that is regarded as false science because it deals with something that has no connection to proper scientific methodology and cannot be proven realistically.

The different types of Pseudosciences include the following:

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Note that the amplitude of the SHM after the collision is 0.275 m. This  is solved using the conservation of momentum and conservation of energy principles.

To solve this problem, we need to use the conservation of momentum and conservation of energy principles.

Before the collision, the velocity of block 1 is in the direction of the spring's length. Therefore, we can assume that block 1 does not affect the SHM of block 2.

Using conservation of momentum:

m1v1 + m2v2 = (m1 + m2)vf

where

m1 = 5.80 kg (mass of block 1)

v1 = 5.70 m/s (velocity of block 1)

m2 = 2.90 kg (mass of block 2)

v2 = Aω (velocity of block 2 in SHM)

vf = (m1v1 + m2v2)/(m1 + m2)

Substituting the values:

(5.80 kg)(5.70 m/s) + (2.90 kg)(Aω) = (5.80 kg + 2.90 kg)vf

Simplifying:

16.63 + 2.9Aω = 8.7vf

Using conservation of energy:

(1/2)kA^2 = (1/2)m2ω^2A^2

where

k = m2ω^2 (spring constant)

A = amplitude of SHM

Substituting the values:

(1/2)(2.90 kg)(2π/0.018 s)^2 A^2 = (1/2)(2.90 kg)(2π/0.018 s)^2 A^2

Simplifying:

kA^2 = m2ω^2A^2

k = m2ω^2

Substituting the values:

(2.90 kg)(2π/0.018 s)^2 = 1307.6 N/m

Now we can solve for A:

(1307.6 N/m)A^2 = (5.80 kg)(5.70 m/s)2 + (2.90 kg)(Aω)2

Substituting the values:

(1307.6 N/m)A^2 = 99.01 J + (2.90 kg)(0.01 m/s)2A^2

Simplifying:

(1307.6 N/m - 0.01 kg/s^2)A^2 = 99.01 J

A^2 = 0.0757 m^2

A = 0.275 m

Therefore, the amplitude of the SHM after the collision is 0.275 m.

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The compression of the spring of spring constant 384 N/m is 2.41 m.

Compression springs work by resisting and pushing back against any downward or inward force that tries to squash and hold them in a compressed state.

To calculate the compression of the spring, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the compression of the spring is 2.41 m.

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The magnitude of the mower's acceleration of mass 36 kg is 1.15 m/s².

Acceleration can be defined as the ratio of the change in velocity to the time of a body.

To calculate the magnitude of the mower's acceleration, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the acceleration of the mower is 1.15 m/s².

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Answer:

185 Newtons

Explanation:

We are solving for the weight of a person on a planet and we are given the mass and acceleration of the person on the planet. Because we know that weight is the force that a person exerts due to gravity and the formula for it is w = mg, all we need to do is plug in our values to get w = (3.7 m/s^2)(50 kg) which = 185 newtons

Answer:

200rad/s

Explanation:

The angular frequency of oscillation of the pendulum is 44.3 rad/s.

Angular frequency is defined as the rate of change of angular displacement in a simple harmonic motion.

Here,

Angular displacement of the pendulum, θ = 0.32 rad

Angular acceleration, α = -630 rad/s²

We know that the equation for angular acceleration is given by,

α = -ω²θ

where ω is the angular frequency

ω² = -α/θ

ω² = 630/0.32

ω² = 1968.8

Therefore,

Angular frequency, ω = √1968.8

ω = 44.3 rad/s

Hence,

The angular frequency of oscillation of the pendulum is 44.3 rad/s.

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At 735m it take for the rocket to hit the ground after it stops moving upward and begins to fall

concept are attached, rest calculation is:

Velocity when the engine is switched off

h max = 2/1 ∗19.6∗5∗5+ 2∗9.8 98∗98

​=245+490=735m

An object is considered to be experiencing uniform acceleration when it is moving straight ahead with an increase in speed that occurs at regular intervals of time. The uniform acceleration of an item during free fall is one example.

When an item travels in a circle at a constant speed, its magnitude of velocity is constant. Its only constant alteration is in direction. It is seen as evenly accelerated motion as a result.

Constant speed is another name for uniform speed. Equal distances can be covered in the same amount of time by a body. Therefore, the body does not accelerate and instead keeps its current pace.

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Explanation:

You want  N/m

 N = 66 * 9.81

 m = 2.3 x 10^-2 m

66* 9.81 / 2.3 x 10^-2  = 28150 =  28 000 N/m    to two S D

Answer:

\(\phi=628.3\)

Explanation:

From the question we are told that

Radius \(r=10.0 cm\)

Magnetic field\(B=2T\)

Generally the equation for area of circular path is mathematically given by

 \(Area=\pi r^2\)

 \(A=\pi 10^2\)

 \(A=314.15m^2\)

Generally the equation for Magnetic flux is mathematically given by

 \(\phi=BA\)

 \(\phi=2*314.15\)

 \(\phi=628.3\)

The position they could possibly have zero electric field is C.

An electric field is a physical field that exists in the space surrounding electrically charged particles or objects. It is a vector field, meaning that it has both magnitude and direction.

When a charged particle is present in a space, it creates an electric field that exerts a force on any other charged particles in the vicinity.

The strength of the electric field at any given point is defined as the force per unit charge that a small test charge would experience if placed at that point.

The magnitude and direction of the electric field depend on the distance between the charges and their magnitudes. The electric field is stronger when the charges are closer together and weaker when they are farther apart.

So position they could experience zero electric field is C, assuming point C if very far from the two charges.

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The net force acting on the future space vehicle be 81750 Newton.

The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

Given parameters:

Mass of the  future space vehicle: m =  250,00 kilograms.

Net acceleration of the   future space vehicle: a = 3.27 m/s^2.

Hence, according to Newton's 2nd law of motion:

Net force acting on the future space vehicle: F = mass × acceleration

= 25000 × 3.27 Newton

= 81750 Newton.

Hence, net force acting on the future space vehicle be 81750 Newton.

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It is to be noted that the magnitude of the electric flux through the hemisphere surface  is πR²E. (Option D)

The characteristic of an electric field known as electric flux may be thought of as the quantity of electric lines of force (or electric field lines) that cross a certain region.

According to this theory, electric field lines begin with positive electric charges and end with negative charges.

Recall  that the A hemispherical surface (half of a spherical surface) of radius R.

In order to derive the Magnitude of the electric flux we use the following formula:

Ф = ∈ x Area of the Curve Surface

⇒ Ф = ∈ x πR²

⇒ Ф = πR²∈

Thus, option D is the correct answer.

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Full Question:

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

A) 2πR²∈

B) 0

C) 4 πR²∈/3

D) πR²∈

Before the ball hits the ground, it is 2.8 seconds in the air. The result is obtained by using the equations in uniformly accelerated straight motion.

The equations apply in uniformly accelerated straight motion in vertical dimension are

v₁ = v₀ + gt

v₁² = v₀² + 2gh

h = v₀t + ½ gt²

Where

A student standing on the ground throws a ball straight up with

Find the time it takes for the ball to reach the ground!

We use g = 9.8 m/s². See the illustration picture in the attachment!

The ball will go upward and stop at a certain height with v₁ = 0. The time needed is

v₁ = v₀ - gt₁

0 = 13.0 - 9.8t

13.0 = 9.8t

t₁ = 13.0/9.8

t₁ = 1.3 s

The height above the hand when the ball stops is

v₁² = v₀² - 2gh₂

0 = 13.0² - 2(9.8)h₂

13.0² = 2(9.8)h₂

169 = 19.6h₂

h₂ = 8,62 m

The ball stops at a height of

h₃ = h₁ + h₂

h₃ = 2.50 + 8.62

h₃ = 11.12 m

The ball goes downward and reach the ground. Initial velocity in this condition is v₁ = 0. The time needed is

h₃ = v₁t + ½ gt₂²

11.12 = 0 + ½ (9.8)t₂²

11.12 = 4.9t²

t₂² = 2.27

t₂ ≈ 1.5 s

The time that the ball in the air is

t = t₁ + t₂

t = 1.3 + 1.5

t = 2.8 s

Hence, the ball is in the air for 2.8 seconds.

Your question is incomplete, but most probably your full question was

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the hand is 2.50 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way).

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a) The average force with which the woman pushed the man is 210 N.

b.  the woman's recoil speed is still 3.68 m/s

c.  If  she now throws her 2.50-kg purse at him at a 20.0° angle, the speed is 0.

We first calculate  the Momentum  = (70.0 kg)(-1.50 m/s) + (55.0 kg)(v) + (2.50 kg)(-1.50 m/s)

0 = (70.0 kg)(0 m/s) + (55.0 kg)(v) + (2.50 kg)(-1.50 m/s) - (70.0 kg)(-1.50 m/s)

v = 3.68 m/s

Δp = (70.0 kg)(1.50 m/s) - (70.0 kg)(0 m/s) = 105 kg·m/s

The average force exerted on the man by the woman is:

Average force =   Δp / Δt

Average force  = 105 kg·m/s / 0.500 s

Average force  = 210 N

b) The woman's recoil speed is :

0 = (70.0 kg)(-1.50 m/s) + (55.0 kg)(v) + (2.50 kg)(-1.50 m/s)

v = 3.68 m/s

c) If the woman throws her 2.50-kg purse at the man at a 20.0° angle, we apply the conservation of momentum equation in two dimensions and get the speed as 0.

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Based on the provided information, we can make the following conclusions that C. The biggest change in cholesterol levels occurred between weeks 2 to 6 of any dose.

It should be noted that the information does not show the specific cholesterol level changes for the 2.5 mg dose. The graph only provides data up to the 40 mg dose.

The biggest change in cholesterol levels occurred between weeks 2 to 6 of any dose: This conclusion is supported by the graph. It shows a significant decrease in cholesterol levels from baseline to week 6 for all the different doses of atorvastatin.

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Based on the provided graph, we can make the following conclusions:

A. 80 mg of atorvastatin lowered cholesterol levels the least:

B. 2.5 mg of atorvastatin lowered cholesterol levels the most:

C. The biggest change in cholesterol levels occurred between weeks 2 to 6 of any dose:

D. The biggest change in cholesterol levels occurred during the first 2 weeks of any dose:

Answer:

  g ≈ 7.4 m/s²

Explanation:

The acceleration due to gravity on planet XX is ...

  g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

  g ≈ 7.4 m/s²

Answer:

one time i was one the flat ground at my aunts house then we went on a hike so i brought my bike it had just rained that day so it was kinda muddy so there was sticks everywhere.i was riding up hill and noticed that it was very hard,then i rode down hill and it was much better

Explanation:

KCl ( potassium chloride ) is classified as solid substance while O₂  ( oxygen molecule ) is classified as gaseous substance.

KCl is an abbreviation for potassium chloride. Potassium chloride ( KCl ) naturally occurs as a white or colorless solid that has a powdery, crystalline appearance.

Potassium chloride ( KCl ) is one of the important raw materials required in the manufacture of potassium metal.

The metal halide salt KCl is also used in the manufacture of soaps.

The chemical formula of potassium chloride ( KCl ), consists of one potassium ( K ) atom and one chlorine ( Cl ) atom.

O₂ is a substance that is known as oxygen molecule and it usually exist in gaseous state.

Thus, we can conclude that we can classify O₂ ( oxygen molecule )which is known as oxygen molecule as gas, and KCl ( potassium chloride ) known as potassium chloride as solid substance.

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