If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).
(a)

Calculate the ideal speed in (m/s) to take an 85 m radius curve banked at 15°.

(b) m/s

What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = \(\frac{W}{g} \ \frac{a}{cos 20}\)

let's calculate

          F = \(\frac{2000}{32} \ \frac{6.6 }{cos20}\)(2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Answer:   the acceleration of the masses is given by = 0, which means the angular acceleration of the pulley is zero. This implies that the masses m and 2m move with constant velocity,  they are in equilibrium.

Answer:

Half

Explanation:

If the half life of an isotope is 20 years, then half of the original amount will  remain after 20 years has passed.

That's what "half-life" means.

Similarly, after ANOTHER 20 years, 1\4 of the original amount remains.

And after ANOTHER 20 years, 1\8 of the original amount remains.

And after ANOTHER 20 years, 1\16 of the original amount remains.

And after ANOTHER 20 years, 1\32 of the original amount remains.

And after ANOTHER 20 years, 1\64 of the original amount remains.

And after ANOTHER 20 years, 1\128 of the original amount remains.

Answer: 25%

Explanation:

From the question, we are informed that the temperature of source and sink of cannot engine are 400K

and 300K respectively.

The efficiency will be calculated as:

= 1 - t/T

= 1 - 300/400

= 1 - 3/4

= 1/4

= 25%

Answer:

\(F_B=235200\ N\)

Explanation:

Given that,

The volume of the cube, V = 24 m³

The density of water, d = 1000 kg/m³

We need to find the buoyant force acting on this cube when it totally submerged in water. The formula for the buoyant force is given by :

\(F_B=dgV\)

Substitute all the values,

\(F_B=1000\times 9.8\times 24\\\\F_B=235200\ N\)

So, the required force is equal to 235200  N.

Answer:

power is 100W

Explanation: power= joule/second

power= 2500/25

=100W

Answer:

To determine the height to which the ball rises before it reaches its peak, we need to know the initial velocity of the ball and the acceleration due to gravity. Let's assume the initial velocity of the ball is v and the acceleration due to gravity is g.

The time it takes for the ball to reach its peak is one-half the total hang-time, or 1/2 * 6.25 s = 3.125 s.

The height to which the ball rises can be calculated using the formula:

height = v * t - (1/2) * g * t^2

Substituting in the values we know, we get:

height = v * 3.125 s - (1/2) * g * (3.125 s)^2

To solve for the height, we need to know the value of v and g. Without more information, it is not possible to determine the height to which the ball rises before it reaches its peak.

Explanation:

Answer:

Approximately \(47.9\; {\rm m}\) (assuming that \(g = 9.81\; {\rm m\cdot s^{-2}}\) and that air resistance on the baseball is negligible.)

Explanation:

If the air resistance on the baseball is negligible, the baseball will reach maximum height at exactly \((1/2)\) the time it is in the air. In this example, that will be \(t = (6.25\; {\rm s}) / (2) = 3.125\; {\rm s}\).

When the baseball is at maximum height, the velocity of the baseball will be \(0\). Let \(v_{f}\) denote the velocity of the baseball after a period of \(t\). After \(t = 3.125\; {\rm s}\), the baseball would reach maximum height with a velocity of \(v_{f} = 0\; {\rm m\cdot s^{-1}}\).

Since air resistance is negligible, the acceleration on the baseball will be constantly \(a = (-g) = (-9.81\; {\rm m\cdot s^{-2}})\).

Let \(v_{i}\) denote the initial velocity of this baseball. The SUVAT equation \(v_{f} = v_{i} + a\, t\) relates these quantities. Rearrange this equation and solve for initial velocity \(v_{i}\):

\(\begin{aligned}v_{i} &= v_{f} - a\, t \\ &= (0\; {\rm m\cdot s^{-1}}) - (-9.81\; {\rm m\cdot s^{-2}})\, (3.125\; {\rm s}) \\ &\approx 30.656\; {\rm m\cdot s^{-1}}\end{aligned}\).

The displacement of an object is the change in the position. Let \(x\) denote the displacement of the baseball when its velocity changed from \(v_{i} = 0\; {\rm m\cdot s^{-1}}\) (at starting point) to \(v_{t} \approx 30.656\; {\rm m\cdot s^{-1}}\) (at max height) in \(t = 3.125\; {\rm s}\). Apply the equation \(x = (1/2)\, (v_{i} + v_{t}) \, t\) to find the displacement of this baseball:

\(\begin{aligned}x &= \frac{1}{2}\, (v_{i} + v_{t})\, t \\ &\approx \frac{1}{2}\, (0\; {\rm m\cdot s^{-1}} + 30.565\; {\rm m\cdot s^{-1}})\, (3.125\; {\rm s}) \\ &\approx 47.9\; {\rm m}\end{aligned}\).

In other words, the position of the baseball changed by approximately \(47.9\; {\rm m}\) from the starting point to the position where the baseball reached maximum height. Hence, the maximum height of this baseball would be approximately \(47.9\; {\rm m}\!\).

Answer:

The Great Pyramid of Giza (also known as the Pyramid of Khufu or the Pyramid of Cheops) is the oldest and largest of the pyramids in the Giza pyramid complex bordering present-day Giza in Greater Cairo, Egypt.It is the oldest of the Seven Wonders of the Ancient World, and the only one to remain largely intact.

Explanation:

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

\(volume = \frac{mass}{density} \\\)

From the question

mass = 2.34 kg

density = 6.45 kg/m³

We have

\(volume = \frac{2.34}{6.45} \\ = 0.36279...\)

We have the final answer as

Hope this helps you

(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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a) 92% Confidence interval: (1027.5,1048.5)

b) Sample size = 100

What is Sample size?

The process of deciding how many observations or replicates to include in a statistical sample is known as sample size determination. Any empirical study with the aim of drawing conclusions about a population from a sample must take into account the sample size as a crucial component.

We are given the following in the question:

Sample mean,  = 1038

Sample size, n = 25

Alpha, α = 0.08

Population standard deviation, σ = 30

a) 92% Confidence interval:

\($\mu \pm z_{\text {critical }} \frac{\sigma}{\sqrt{n}}$\)

Putting the values, we get,

\($z_{\text {critical }}$\) at \($\alpha_{0.08}=\pm 1.75$\)

\($1038 \pm 1.75\left(\frac{30}{\sqrt{25}}\right)=1038 \pm 10.5=(1027.5,1048.5)$\)

b) In order to reduce the confidence interval by half, we have to quadruple the sample size.

Thus,

Sample size \($=25 \times 4=100$\)

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Answer:

40 J

Explanation:

\(KE = \frac{1}{2} mv^{2} \\KE = \frac{1}{2} (0.200 kg)(20 m/s)^{2} \\KE = 40 J\)

If the dog walked at a constant speed the whole way, the graph of the dog's position versus time would be a straight line. This is because the dog's velocity (which is the derivative of position with respect to time) would be constant, and the acceleration (which is the derivative of velocity with respect to time) would be zero.

A straight line on a position-time graph indicates that the object is moving at a constant velocity. The slope of the line would be equal to the velocity of the dog.

If the graph is a horizontal line, it would indicate that the dog is at rest. If the line slopes upward, the dog is moving in the positive direction (for example, to the right in a position-time graph), and if the line slopes downward, the dog is moving in the negative direction.

In all, A constant speed means a constant velocity and the line is a straight line with a particular slope.

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The net work done on the cart is 1.17 J.

The average net force on the cart is 1.65 N.

The work done by frictional force over the same distance is 0.078 J.

The net work done on the cart is calculated by applying the principle of conservation of energy as follows;

W  = ΔK.E

W = ¹/₂m(v² - u²)

where;

W = ¹/₂(0.53)(2.1² - 0²)

W = 1.17 J

The average net force on the cart is calculated as follows;

W = Fd

where;

F = W/d

F = (1.17) / (0.71)

F = 1.65 N

The work done by frictional force over the same distance is calculated as follows;

W = fd

W = 0.11 x 0.71

W = 0.078 J

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The impulse exerted by Earth on the apple during the first 0.50 m of its fall is 0.74 Ns, and during the next 0.50 m, it is 0.37 Ns.

Using the equation for impulse, which is impulse = force x time, we can calculate the impulse that Earth exerts on the apple during the first 0.50 m and the next 0.50 m of its fall.

First, we need to calculate the force of gravity acting on the apple, which is given by the equation F = mg, where m is the mass of the apple and g is the acceleration due to gravity (approximately 9.81 m/s^2).

The mass of the apple is 150 g, which is 0.15 kg. Therefore, the force of gravity acting on the apple is:

F = mg = (0.15 kg)(9.81 m/s^2) = 1.47 N

Now, we can calculate the impulse exerted by Earth on the apple during the first 0.50 m of its fall. Since the force of gravity is constant, we can use the equation impulse = force x distance, where distance is the distance over which the force is applied.

Impulse during first 0.50 m = force x distance = (1.47 N)(0.50 m) = 0.74 Ns

For the next 0.50 m of the apple's fall, we need to consider that the velocity of the apple is increasing, so the force of gravity is no longer constant. However, we can approximate the average force over this distance as half the force at the start of the fall, or 0.5(1.47 N) = 0.74 N.

Using the same equation impulse = force x distance, we can calculate the impulse exerted by Earth on the apple during the next 0.50 m of its fall:

Impulse during next 0.50 m = force x distance = (0.74 N)(0.50 m) = 0.37 Ns.

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Answer:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

Explanation:

A dipole is a pair of magnetized, equal or oppositely charged poles the are being separated by a distance.

The statements about a dipole that are correct are:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

Answer:

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s

Answer:

the answer is true

Explanation:

Answer:

Explanation:

We can use the conservation of energy principle to find the force constant of the spring. Initially, the ball has potential energy due to its height above the spring, and no kinetic energy. When it strikes the spring, it compresses it, and its potential energy is converted to elastic potential energy in the spring. At the instant the ball comes to rest, all of its initial potential energy has been converted to elastic potential energy in the spring.

The potential energy of the ball at height h is given by:

U = mgh

Where m is the mass of the ball, g is the acceleration due to gravity, and h is the initial height of the ball.

The elastic potential energy stored in the spring when it is compressed by distance d is given by:

U = (1/2)kx^2

where k is the force constant of the spring, and x is the compression distance.

Since the potential energy of the ball is converted entirely to elastic potential energy in the spring, we can set these two expressions equal to each other:

mgh = (1/2)kx^2

Plugging in the given values, we have:

m = 1.55 kg

g = 9.81 m/s^2

h = 0.68 m

d = 0.091 m

Substituting these values into the equation, we get:

(1.55 kg)(9.81 m/s^2)(0.68 m) = (1/2)k(0.091 m)^2

Solving for k, we get:

k = (2mgh)/(d^2) = 2365 N/m

Therefore, the force constant of the spring is 2365 N/m.

The statement that is true about the theory of plate tectonics and the theory of continental drift C. The theory of plate tectonics gives the method by which continents can move as part of plates .

According to the scientific hypothesis of plate tectonics, the underground movements of the Earth create the primary landforms. By explaining a wide range of phenomena, including as mountain-building events, volcanoes, and earthquakes, the theory, which became firmly established in the 1960s, revolutionized the earth sciences.

The scientist Alfred Wegener is most closely connected with the concept of continental drift. Wegener wrote a paper outlining his notion that the continents were "drifting" across the Earth, occasionally crashing through oceans and into one another, in the early 20th century.

According to tectonic theory, the Earth's surface is dynamic and can move up to 1-2 inches every year. The numerous tectonic plates constantly move and interact. The outer layer of the Earth is altered by this motion. The result is earthquakes, volcanoes, and mountains.

Therefore, option C is correct.

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Answer:

\(Vf^2=Vo^2+2aS\\(0m/s)^2=(4.67m/s)^2+(2*-10m/s^2)S\\-(4.67)^2 m^2/s^2=-20m/s^2*S\\S=(21.8089/20) m\\S=1.090445 m\\\)

With the use of third equation of motion, the sled be moving as fast as 40.24 m/s when it reaches the other end of the field.

Acceleration is the rate at which velocity is changing. It is a vector quantity.

Given that you  accelerates at a rate of 3.95 m/s2 when the rocket is on across a field of 205 m. To know how fast the sled will move when it reaches the other end of the field, we will use third equation of motion. That is, v² = u² + 2as

Where

Substitute all the parameters into the formula

v² = 0 ² + 2 × 3.95 × 205

v² = 1619.5

v = √1619.5

v = 40.24 m/s

Therefore, the sled be moving as fast as 40.24 m/s when it reaches the other end of the field.

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Answer:

Gravitional pull or gravity

Explanation:

Its that simple

Answer:

Measurement is the comparison of any physical quantity of an object to a standard unit which is pre-determined. The standard units such as length, time, mass etc are known as the Fundamental units of Measurement.

Explanation:

Any object that can be measured is known as a physical quantity. So, to measure the physical quantity, we require some standard units. A measurement consists of two parts - the numerical measurement and the standard unit which is pre-determined. For example, the length of a given table is 10cm, which implies that 10 is the numerical value and the standard unit of measurement is centimeter (cm).

Measurements can be both Fundamental and Derived. Examples of Fundamental quantities are Length, Time etc, while example of Derived quantity is speed which is derived from Length and Time.

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Answer:

Part A

You have to ride 8.0 km before turning north to get to your friend’s house.

Part B

The sine of the angle  θ at the location of the friend's house is 0.8

Explanation:

The remaining part of the question which is an image is attached below

Explanation:

Part A

To determine how far you will ride ride before turning north,

From the diagram, that is the distance of your street.

Let the distance of your street be \(A\)

and the distance of your friend's street be \(B\)

and let the displacement between your friends house and your house be \(C\)

The relation in the diagram shows a right angle triangle.

The sides of the right angle triangle are represented as \(A,B\) and \(C\).

To find \(A\), which is the distance of your street,

From Pythagorean theorem, 'The square of hypotenuse is the sum of squares of the other two sides'

That is,

\(/Hypoyenuse/^{2} = /Adjacent/^{2} + /Opposite/^{2}\)

\(C\) is the hypotenuse, which is the displacement between your friends house and your house,

Hence, \(C = 10.0 km\)

\(B\) is adjacent, which is the distance of your friends street

then, \(B = 6.0 km\)

and \(A\) is the opposite, which is the distance of your house

From Pythagoras theorem, we can then write that,

\(C^{2} = B^{2} + A^{2}\)

Then, \(10.0^{2} = 6.0^{2} + A^{2}\)

\(A^{2} = 100.0 - 36.0\\A^{2} = 64.0\\A = \sqrt{64.0}\)

\(A = 8.0km\)

Hence, you have to ride 8.0 km before turning north to get to your friend’s house.

Part B

To find the sine of the angle  θ at the location of the friend's house,

In the diagram, the sine of the angle  θ is given by

\(Sin\theta = \frac{Opposite}{Hypotenuse}\)

Hence, \(Sin\theta = \frac{A}{C}\)

Then,

\(Sin\theta = \frac{8.0}{10}\)

\(Sin\theta = 0.8\)

Hence, the sine of the angle  θ at the location of the friend's house is 0.8

A. The amount of distance you have to ride before turning North to get to your friend’s house is 8 kilometers.

B. The sine of the angle (θ) at the location of your friend's house is 0.8.

Given the following data:

A. To determine the amount of distance you have to ride before turning North to get to your friend’s house, we would apply Pythagorean's theorem:

Mathematically, Pythagorean's theorem is given by the formula:

\(c^2 = a^2 + b^2\\\\10^2 =6^2+b^2\\\\100=36+b^2\\\\b^2 =100-36\\\\b^2 =64\\\\b=\sqrt{64}\)

b = 8 kilometers

B. To find the sine of the angle (θ) at the location of the friend's house:

Mathematically, the sine of an angle is given by the formula:

\(Sin\theta = \frac{opposite}{hypotenuse}\)

Substituting the given parameters into the formula, we have;

\(Sin\theta = \frac{8}{10} \\\\Sin\theta = 0.8\)

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Answer:

1 rev = 2(pi) rad  pi(rad) = 180 degrees

so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees

Explanation:

63.36 estimated to 63 so 63

The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

1 rev = 2(pi) rad

So here  pi(rad) = 180 degrees

Now for 33 rpm it should be like

=  33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s

= 63.36 degrees

= 63 degrees

hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

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Answer:

North and South poles

Explanation:

The speed of the artillery shell when it hits its target is 100 m/s.

Given:

Initial vertical displacement (y) = 200 m

Vertical displacement at 100 m in the air (y') = 100 m

Final velocity in the vertical direction (vy') = 0 m/s (at the highest point of the trajectory)

Using the equation for vertical displacement in projectile motion:

y' = vy^2 / (2g),

where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can solve for the initial vertical velocity (vy).

100 m = vy^2 / (2 * 9.8 m/s^2),

vy^2 = 100 m * 2 * 9.8 m/s^2,

vy^2 = 1960 m^2/s^2,

vy = sqrt(1960) m/s,

vy ≈ 44.27 m/s.

Now, since the horizontal motion is independent of the vertical motion, the horizontal speed of the shell remains constant throughout its trajectory. Therefore, the speed of the shell when it hits its target is 100 m/s.

Hence, the speed of the artillery shell when it hits its target is 100 m/s.

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Answer:

7 because salt lake and Southis weat