a canoe that weighs 600 n floats in a lake. what is the buoyant force on the canoe

Answer:

position 1

Explanation:

in position 2 and three it would have less energy cause there is less force being exerted to give it that push, but at the end position and the first position there is going to be more force and energy behind the ball.

To calculate the total amount of money Anna will have at the end of four years, we need to consider the regular deposits she makes and the interest earned on those deposits.

First, let's determine the number of deposits Anna will make over four years. Since she deposits $150 every other month, there are 12 months in a year, and she makes deposits every two months, she will make a total of 12/2 = 6 deposits per year. Therefore, over four years, she will make 4 * 6 = 24 deposits.

Next, let's calculate the interest earned on these deposits. The annual interest rate is 3.5%, compounded bimonthly. To calculate the bimonthly interest rate, we divide the annual interest rate by the number of compounding periods in a year, which is 12/2 = 6. So, the bimonthly interest rate is 3.5% / 6 = 0.5833%.

Using the formula for compound interest, A = P(1 + r/n)^(nt), where A is the future value, P is the principal (deposit amount), r is the interest rate per period, n is the number of compounding periods per year, and t is the number of years, we can calculate the future value.

Let's plug in the values:

P = $150

r = 0.5833% (0.005833 as a decimal)

n = 6

t = 4

A = 150(1 + 0.005833/6)^(6*4)

Calculating this expression will give us the total amount of money Anna will have at the end of four years.

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Answer:

The description as per the given scenario is described below throughout the segment.

Explanation:

Answer:

impulse = force x time

impulse = 1900000 x 0.005 = 9500Ns

Answer: The answer is C.

Explanation: 1 centimeter is equal to 0.01 of a meter and 178 centimeters times 0.01 is 1.78 meters, Hope that helps

The image will form at the focal point

1/object + 1/image = 1/focal

If the object distance is infinite then the image distance equals the focal distance

In the event of an accident or a car crash, road safety barriers and fences prevent automobiles from running off the road.

Newton's Three Laws of Physics can help explain what these safety barriers are and how they help to save lives during car accidents :

I.  Unless acted upon by an imbalanced force, an object at rest will remain at rest, and an object at constant velocity will remain at constant velocity.

II.  If an imbalanced force occurs, a mass will experience acceleration proportional to its magnitude.

III.  When you apply a force to an object, you will feel a force that is equal    in magnitude but opposite in direction.

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The forces exerted on the ball when the ruler touches it will depend on a variety of factors such as the material of the ball and ruler, the speed and angle of contact, and the mass of the ball and ruler.

Assuming that the ruler and ball are made of solid materials and the ruler is touching the ball with a small force, the primary forces that will be exerted on the ball are:

Normal force: This is the force exerted by the ruler on the ball perpendicular to the surface of contact. It is equal in magnitude and opposite in direction to the force exerted by the ball on the ruler, according to Newton's third law.

Frictional force: This is the force that opposes the motion of the ball relative to the ruler. It acts tangentially to the surface of contact and is proportional to the normal force and the coefficient of friction between the two surfaces.

Air resistance: This is the force exerted by the air molecules on the ball as it moves through the air. It acts in the opposite direction to the velocity of the ball and is proportional to the velocity and surface area of the ball.

Gravity: This is the force that pulls the ball towards the center of the Earth. It acts vertically downwards and is proportional to the mass of the ball and the acceleration due to gravity.

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If the isotope has a half-life of 1150 years, this means that every 1150 years the amount of the isotope is halved. After one half-life, the amount is reduced to 1/2, after two half-lives it is reduced to 1/4, after three half-lives it is reduced to 1/8, and so on.

To determine how many years are required for the isotope to decay to 25% of its original amount, we need to determine how many half-lives it takes to get from 100% to 25%.

25% is the same as 1/4, so we need to determine how many times we need to halve the original amount to get to 1/4.

1/4 = (1/2)^n, where n is the number of half-lives

Solving for n:

n = log(1/4) / log(1/2)

n = 2

This means that it takes two half-lives for the isotope to decay to 25% of its original amount.

Since the half-life is approximately 1150 years, the time required for two half-lives is approximately:

2 x 1150 years = 2300 years

Therefore, it would take approximately 2300 years for a given amount of this isotope to decay to 25% of that amount.

The Kelvin scale is known as the absolute scale. The lowest temperature in Kelvin is the temperature 0 (there is no negative temperature in Kelvin).

The conversion from Kelvin to Fahrenheit is given by the formula below:

Using K = 0 in this formula, we have:

So the lowest temperature in Fahrenheit is -459.67 °F.

The reason galaxies that are distant from our galaxy move away from our galaxy more rapidly is more space expands between us and distant galaxies.

A galaxy is a group of millions of stars and their systems that are grouped due to gravitational forces.

According to the Big Bang theory, galaxies are expanding and separate among them.

In conclusion, the reason galaxies that are distant from our galaxy move away from our galaxy more rapidly is more space expands between us and distant galaxies.

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Answer:

the galaxies are moving as the universe keep expanding .

Explanation:

Answer:

Roughly 4.5 billion years

The dot product is 25, the angle is \(\theta = cos^{-1} \frac{25}{\sqrt{35} \times \sqrt{21}}\), the cross product is 1i + (-6)j + (-7)k, and the area of the parallelogram spanned by vectors A and B is \(\sqrt{86}\).

Given,

A = 5i + 3j + k

B = 4i + j + 2k

i. Dot Product (A · B):

The dot product of two vectors A and B is given by the sum of the products of their corresponding components.

\(A.B = (A_x \times B_x) + (A_y \times B_y) + (A_z \times B_z)\\A.B = (5 \times 4) + (3 \times 1) + (1 \times 2) \\= 20 + 3 + 2 \\= 25\)

ii. Angle between vectors A and B:

The angle between two vectors A and B can be calculated using the dot product and the magnitudes of the vectors.

\(cos\theta = (A.B) / (|A| \times |B|)\\\theta = \frac{1}{cos} ((A.B) / (|A| \times |B|))\\A = \sqrt{(5^2 + 3^2 + 1^2)} =\\ \sqrt{35}\\B = \sqrt{(4^2 + 1^2 + 2^2)} \\= \sqrt{21}cos\theta = \frac{(A.B) / (|A| \times |B|)\\\theta = \frac{1}{cos} \frac{25}{\sqrt{35} \times \sqrt{21}}}\)

iv. Cross Product (A × B):

The cross product of two vectors A and B is a vector that is perpendicular to both A and B and its magnitude is equal to the area of the parallelogram spanned by A and B.

\(A\times B = (A_y \timesB_z - A_z \timesB_y)i + (A_z \timesB_x - A_x \timesB_z)j + (A_x \times B_y - A_y \times B_x)k\\A\times B = ((3 \times 2) - (1 \times 1))i + ((1 \times 4) - (5 \times 2))j + ((5 \times 1) - (3 \times 4))k\\= 1i + (-6)j + (-7)k\)

Area of the parallelogram spanned by vectors A and B:

The magnitude of the cross product A × B gives us the area of the parallelogram spanned by A and B.

Area = |A × B|

Area of the parallelogram spanned by vectors A and B:

Area = |A × B| =

\(\sqrt{(1^2 + (-6)^2 + (-7)^2}\\\sqrt{1+36+49\\\\\sqrt{86}\)

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This result shows that the absolute derivative of a contravariant vector field Va along a curve γ in a coordinate neighborhood U is equal to the covariant derivative of Va along γ.

In order to show that the absolute derivative of a contravariant vector field Va along a curve γ is equal to the covariant derivative of the vector field Va along γ, we can start by using the definition of the absolute derivative.

The absolute derivative of a contravariant vector field Va along a curve γ is given by:

DVa/dτ = dVa/dτ + Γa_{bc} * V^b * dγ^c/dτ,

where Va represents the contravariant vector field, d/dτ denotes the derivative with respect to the curve parameter τ, Γa_{bc} are the Christoffel symbols of the second kind, V^b represents the components of the vector field Va, and dγ^c/dτ represents the components of the tangent vector to the curve γ.

Now, let's consider the expression DVa/dτ. Since we are dealing with a coordinate neighborhood U in a manifold M, the Christoffel symbols can be expressed in terms of the metric tensor g_{ab} and its derivatives:

Γa_{bc} = (1/2) * g^{ad} * ( ∂g_{bd}/∂x^c + ∂g_{cd}/∂x^b - ∂g_{bc}/∂x^d ),

where g^{ad} represents the contravariant components of the inverse metric tensor.

Substituting this expression into the absolute derivative formula, we have:

DVa/dτ = dVa/dτ + (1/2) * g^{ad} * ( ∂g_{bd}/∂x^c + ∂g_{cd}/∂x^b - ∂g_{bc}/∂x^d ) * V^b * dγ^c/dτ.

Now, let's consider the covariant derivative of the vector field Va along the curve γ, denoted by ∇_V Va. The covariant derivative is defined as:

∇V Va = dVa/dτ + Γa{bc} * V^b * dγ^c/dτ.

Comparing this with the expression for the absolute derivative, we can see that they are identical. Therefore, we have:

DVa/dτ = ∇_V Va.

This result shows that the absolute derivative of a contravariant vector field Va along a curve γ in a coordinate neighborhood U is equal to the covariant derivative of Va along γ.

Note: It's important to mention that the above derivation assumes a Riemannian or pseudo-Riemannian manifold with a metric tensor. The specific form of the Christoffel symbols depends on the metric tensor and the coordinate system used.

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Given a coordinate neighborhood U in a manifold M i.e. U ⊆ M. Let γ be a curve in U.

Using the definition of the absolute derivative of a contravariant vector field Va, we need to show that DVα/dτ = 0 for all α in Va.

Using the definition of the absolute derivative, we have

DVα/dτ = (d/dτ)(αβ ∂/∂xβ) = (dαβ/dτ) ∂/∂xβ + αβ(d/dτ)(∂/∂xβ) …(1)

The partial derivative of ∂/∂xβ along γ is given by

(1/C)(∂/∂xβ)γC …(2)

where C(τ) is the coordinate representation of γ.Using (2) in (1), we get

DVα/dτ = (dαβ/dτ) ∂/∂xβ + (αβ/C)(d/dτ)(∂/∂xβ)γC …(3)

We know that(d/dτ)(∂/∂xβ)γC = (∂/∂xβ)(dγC/dτ) …(4)

We also know that γ(τ) ∈ U, which means that the coordinate functions xβ(γ(τ)) are smooth on the domain of γ. Hence, by definition,

∂xβ/∂τ = d(xβ ◦ γ)/dτ = (d/dτ)xβ(γ(τ)) …(5)

Using (5) in (4), we get

(d/dτ)(∂/∂xβ)γC = (d/dτ)xβ(γ(τ)) (∂/∂xβ)γC = ∂xβ/∂τ (∂/∂xβ)γC …(6)

Using (6) in (3), we get

DVα/dτ = (dαβ/dτ) ∂/∂xβ + (αβ/C)∂xβ/∂τ (∂/∂xβ)γC …(7)

Since xβ(γ(τ)) are smooth functions, ∂xβ/∂τ and (∂/∂xβ)γC are both smooth. Hence, the second term on the right-hand side of (7) is a smooth vector field along γ. It remains to show that the first term on the right-hand side of (7) is also a smooth vector field along γ. This is true since αβ is a smooth tensor field on U, and dαβ/dτ is obtained by differentiating αβ along γ, which preserves smoothness.

Hence, DVα/dτ is a smooth vector field along γ, and its coordinate expression is

DVα/dτ = (dαβ/dτ) ∂/∂xβ …(8)

Using (8), we get

DVα/dτ = 0 for all α in Va.

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Answer:

Please translate this into Spanish! Sorry I can't :(

Explanation:

Disadvantages of Friction:

Friction produces unnecessary heat leading to the wastage of energy.

The force of friction acts in the opposite direction of motion, so friction slows down the motion of moving objects.

Forest fires are caused due to the friction between tree branches.

The magnitude of a uniform electric field is 1.47 × \(10^{8}\) V/m.

What is an electric field?

When charge is present in any form, a point in space has an electric field that is connected to it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. Without any precise information of what generated the field, simply knowing the value of the electric field at a certain location is sufficient to predict what would happen to electric charges nearby.

Given,

B = 0.49 T

Energy density due to the magnetic field

\(u_{b}\) = \(\frac{1}{2mu_{0}}\)B²

Energy density due to the electric field

\(u_{e}\) = \(\frac{1}{2}\)ε₀E²

We know, \(u_{b}\) = \(u_{e}\)

\(\frac{1}{2mu_{0} }\)B² = \(\frac{1}{2}\)ε₀E²

\(\frac{1}{mu_{0} }\)B² = ε₀E²

E² = \(\frac{1}{mu_{0}E_{0} }\)B²

E =  \(\frac{1}{\sqrt{mu_{0}E_{0}}}\)B

\(\frac{1}{\sqrt{4\pi X 10^{-7} X 8.852 X 10^{-12}}}\) × 0.49

3 × \(10^{8}\) × 0.49

= 1.47 × \(10^{8}\) V/m

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Answer: \(F_4\)

Explanation:

The friction opposes motion. We can assume the motion to be downwards because the only force is the weight of the box mgsin(theta) aka F3.

Arrow 4 correctly indicates the direction of friction acting on the box.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.

Friction is a force that resists motion. Because the only force acting on the box is its own weight, we may conclude it is moving downhill. As the force approach the downhill the friction force is acting opposite to it.

Hence, arrow 4 correctly indicates the direction of friction acting on the box.

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Explanation:

Fourteen yard (14) yard

Answer:

C. 157 bar/2270 psi

Explanation:

Calculation to determine what we should head back when either of our SPGs read

SPGs=200 bar -[200 bar-(50 bar + 20 bar)]÷1/3]

SPGs=200 bar-[(200 bar-70 bar)÷1/3]

SPGs=200 bar-(130 bar÷1/3)

SPGs=200 bar-43 bar

SPGs=157 bar/2270 psi

Therefore based on the above calculation we should head back when either of our SPGs read 157 bar/2270 psi

41.38 was the cat's vertical height in the tree .

To find the cat's vertical height in the tree, we use the trigonometric function tangent, which is given by:

tan θ = perpendicular/base

Where,θ = 62 degrees

base = 22 feet

Let perpendicular height of the cat in the tree = x feet

From the given data, we have to find the vertical height of the cat in the tree. The tangent of an angle is equal to the length of the opposite side of the right angle triangle divided by the adjacent side of the right angle triangle.

Thus we can form an equation as follows:

tan 62 = x/22

Solving this equation for x we have;

x = 22 tan 62= 22(1.881)≈ 41.38 feet.

Therefore, the cat's vertical height in the tree is approximately 41.38 feet.

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Answer:

C.endothermic

Explanation:

chemical reactions

Answer:

C

Explanation:

The total work done on the crate of bananas is 4800 J (the work done to move it horizontally).

Work done is defined as the product of displacement and force, w = FS.

Here,

There are two parts to the work done on the crate of bananas: the work done to move it horizontally and the work done to lift it vertically.

First, let's calculate the work done to move the crate horizontally. The formula for work is,

Work = Force x Distance x cos(Ф)

where theta is the angle between the force and the displacement. In this case, the force is the tension in the rope, and it is pulling horizontally. So, theta is 0 degrees, and cos(Ф) is 1.

The distance the crate is moved horizontally is 5 meters, and the force applied by the motor is,

Force = Power / Velocity = 40 W / (5 m / 120 s) = 960 N

So, the work done to move the crate horizontally is:

Work = 960 N x 5 m x cos(0) = 4800 J

The distance the crate is lifted vertically is not given in the problem, so we cannot calculate the work done to lift the crate without this information.

Therefore, the total work done on the crate of bananas is 4800 J (the work done to move it horizontally).

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Answer:

Canidate B

Explanation:

Becuase he/she is good to work with, and will most likely be more determined.

The final speed of the sled was approximately 243 miles per hour.

The question requires us to calculate the final speed of the sled accelerated over a distance of 15.40 km by rockets that supplied a constant force of 23.310 Newtons. The sled weighed a total of 192 pounds and ignoring friction/air resistance. We can use the formula of kinematics to find the final speed of the sled. That is,

Final velocity (v) = Initial velocity (u) + (Acceleration (a) x Time (t))

From Newton's Second Law, F = ma

Where m is the mass of the sled and a is the acceleration produced by the force F. Therefore, acceleration produced by force F can be given by the formula,

a = F/mPutting values of F and m in the above formula, we have,

a = (23.310 N) / (192 lbs) x (1 lb / 4.448 N) = 0.02764 m/s²The given distance is in km. We need to convert it into meters as the unit of acceleration is m/s².

Distance = 15.40 km = 15.40 × 1000 m = 15,400 m

Initial velocity, u = 0 (as the sled was initially at rest)

Therefore, using the above kinematic formula, we get,v = (2 × a × s)1/2

Where s is the distance traveled by the sled.

v = (2 × 0.02764 m/s² × 15400 m)1/2 = 108.7 m/s

Converting m/s into miles/hour108.7 m/s = 108.7 x 2.237 = 243 miles/hour

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Two vertical wires are used to hoist a 12 kilogram lantern from the ceiling; as a result, each wire is under 58.8 N of tension.

The phrase "wire transfer" dates back to a time when banks used telegraph wires to transmit this type of financial transfer. A bank wire stored procedures on who should receive the funds, along with the user's bank account information and the amount they should receive.

Well, we know from Newton that, if m is the mass of the lantern,

Fnet = ma

Additionally, it is clear that the two forces pulling on the candle are now the strain on the two wires and the gravitational force. Since the candle isn't moving, its speed is 0, which means that the final quantity of forces acting on it must likewise be zero.

The positive y-axis should point up, and also the negative y-axis should point below. Listing the forces reveals that

-mg + T1 + T2 = 0

The two wires' tensions are also equal because they are both straight and orbital angular momentum, or T1 = T2 = T. The preceding equation can be revised to say

-mg + 2T = 0, or

2T = mg.  

Thus,

T = mg/2 = (12kg*9.8m/s2) / 2 = 58.8N

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