The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it is turning at 4.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 11.0 s. Through how many revolutions does the tub rotate while it is in motion?

Answers

Answer 1

Answer:

36 rev

Explanation:

See it in the pic

The Tub Of A Washing Machine Goes Into Its Spin Cycle, Starting From Rest And Gaining Angular Speed Steadily

Related Questions

A mail carrier parks her postal truck and delivers packages. To do so, she walks east at a speed of 0.80 m/s for 4.0 min, then north at a speed of 0.50 m/s for 5.5 min, and finally west at a speed of 1.1 m/s for 2.8 min. Define east as +x and north as +y. (a) Write unit-vector velocities for each of the legs of her journey. (b) Find unit-vector displacements for each of the legs of her journey. (c) Find her net displacement from the postal truck after her journey is complete.

Answers

Explanation:

Given that,

She walks in east,

Speed = 0.80 m/s

Time = 4.0 min

In north,

Speed = 0.50 m/s

Time = 5.5 min

In west,

Speed = 1.1 m/s

Time = 2.8 min

(a). We need to calculate the unit-vector velocities for each of the legs of her journey.

The velocity of her in east

[tex]\vec{v_{1}}=0.80\ \hat{x}\ m/s[/tex]

[tex]\vec{v_{2}}=0.50\ \hat{y}\ m/s[/tex]

[tex]\vec{v_{3}}=1.1\ \hat{-x}\ m/s[/tex]

(b). We need to calculate the unit-vector displacements for each of the legs of her journey

Using formula of displacement

[tex]\vec{d_{1}}=v_{1}\times t_{1}[/tex]

In east ,

[tex]\vec{d_{1}}=0.80\times4.0\times60[/tex]

[tex]\vec{d_{1}}=192\ \hat{x}\ m[/tex]

In north,

[tex]\vec{d_{2}}=0.50\times5.5\times60[/tex]

[tex]\vec{d_{2}}=165\ \hat{y}\ m[/tex]

In west,

[tex]\vec{d_{3}}=1.1\times2.8\times60[/tex]

[tex]\vec{d_{3}}=184.8\ \hat{-x}\ m[/tex]

(c). We need to calculate the  net displacement from the postal truck after her journey is complete

[tex]\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}[/tex]

Put the value in the formula

[tex]\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}[/tex]

[tex]\vec{d}=7.2\hat{x}+165\hat{y}[/tex]

We need to calculate the magnitude of the displacement

[tex]d=\sqrt{(7.2)^2+(165)^2}[/tex]

[tex]d=165.16\ m[/tex]

The magnitude of the displacement is 165.16 m.

Hence, This is the required solution.

The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuck in the sandy bottom of the pond. If the stick is 19 cm tall, how long is the shadow?

Answers

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

[tex]n_{air} sin30 =n_{water} sin\theta[/tex]

[tex]1*0.5 = 1.33*sin\theta[/tex]

solving for[tex] \theta[/tex]

[tex]sin\theta = \frac{3}{8}[/tex]

[tex]\theta = sin^{-1}\frac{3}{8}[/tex]

[tex]\theta =  22 degree[/tex]

from shadow- stick traingle

[tex]tan(90-\theta) = cot\theta = \frac{h}{s}[/tex]

[tex]s = \frac{h}{cot\theta} = h tan\theta[/tex]

s = 19tan22 = 7.67 cm

s = shadow length

The calculated shadow length is approximately 10.97 cm.

To determine the length of the shadow cast by the stick, we can use basic trigonometry.

Specifically, we'll use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.

Given:

The height of the stick (opposite side) is 19 cm.The angle of elevation of the sun is 60°.

The formula for tangent is:

tan(θ) = opposite / adjacent

Here, the opposite side is the height of the stick, and the adjacent side is the length of the shadow. Thus, we can rearrange this formula to solve for the length of the shadow (adjacent side):

adjacent = opposite / tan(θ)

Substituting the given values:

adjacent = 19 cm / tan(60°)

We know that tan(60°) is √3 or approximately 1.732.

So:

adjacent = 19 cm / 1.732 ≈ 10.97 cm

Therefore, the length of the shadow cast by the stick is approximately 10.97 cm.

The fastest measured pitched baseball left the pitcher's hand at a speed of 42.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

Answers

Answer:

The acceleration and time are 588 m/s² and 0.071 sec respectively.

Explanation:

Given that,

Speed = 42.0 m/s

Distance = 1.50 m

(a). We need to calculate the acceleration

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

Put the value in the equation

[tex]a=\dfrac{42.0^2-0}{2\times1.50}[/tex]

[tex]a=588\ m/s^2[/tex]

(b). We need to calculate the time

Using equation of motion

[tex]v = u+at[/tex]

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{42.0-0}{588}[/tex]

[tex]t=0.071\ sec[/tex]

Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.

You are jogging for one hour at a speed of 5 m/s and for the final 200 m you decide to increase your velocity to 11 m/s. Assuming you were able to do so instantaneously. Once you reach that speed how long it will take to cover 200m?

Answers

Answer:

[tex]t=18.18s[/tex]

Explanation:

From the exercise we know that the person change its initial velocity from 5m/s to 11 m/s for 200m

According to the formula:

[tex]v=\frac{x}{t}[/tex]

If we want to know how much time does it take to cover 200 m at 11 m/s we need to calculate the following formula:

[tex]t=\frac{x}{v}=\frac{200m}{11m/s}=18.18s[/tex]

Zeta Anderson, futuristic super-spy for the Terran Confederation, has completed her objective of stealing intelligence from the Zorn collective. Stealthily, she slips into her space suit (with jet-pack), and slips from an airlock, headed for her stealth ship. Her jet-pack can supply her with a constant acceleration, and gravity can be neglected. When she turns on her jet-pack, how does her velocity change? Since she does not want to be going too fast (and either overshoot, or collide with her ship), how does her velocity change when she turns the jet-pack off?

Answers

Explanation:

First lets understand the 2nd law of motion by Sir Isaac Newton. According to this law,

[tex]Force = mass\times acceleration[/tex]

[tex]Velocity = acceleration\times time[/tex]

Since the spy is in space there is no medium and hence no friction to restrict the motion. Thus, when the spy turns on the jet pack, she will be accelerated and her velocity will increase. As there is no medium so no friction. So even when she turns her jet-pack off the velocity will not change. Although the acceleration will be zero but she will be moving with a constant velocity until an opposite force is applied. That can be done using reverse thrust.

An object (with charge -2.4 µC and mass 0.026 kg) hovers at rest above the ground, its weight being held up by a uniform electric field. (a) Find the size and direction of this electric field.
(b) If the electric charge on the object is quadrupled while its mass remains the same, find the size and direction of its acceleration.

Answers

Answer:

a ) 10.61 X 10⁴ N/C

b )29.37 ms⁻² .

Explanation:

Since the weight of the object is balanced by force by electric field , the force must be acting in upward direction . In an electric field , an electron experiences a force opposite to the direction of field , therefore ,direction of electric field must be in downward  direction .

Let E be the electric field . Force on the charge

= electric field x charge = E x 2.4 x 10⁻⁶

Weight acting downwards = .026 x 9.8

For balancing

.026 x 9.8 =  E x 2.4 x 10⁻⁶

E = .026 x 9.8 / 2.4 x 10⁻⁶

=  10.61 X 10⁴ N/C

b )

If the charge is quadrupled , charge becomes 4 x 2.4 x 10⁻⁶

Upward force = 4 x 2.4 x 10⁻⁶ x 10.61 X 10⁴ = 101.85 x 10⁻²

Down ward force =.026 x 9.8

Net force = 101.85 x 10⁻² - .026 x 9.8

= .7637 N

acceleration = .7637 / .026 = 29.37 ms⁻² .

What is the significance of the direction of an electric field line at some point on the line?

Answers

Answer with Explanation:

The direction of the electric field line at any point gives us the direction of the electric force that will act on a positive charge if placed at the point. We know that if we place a charge in an electric field it will experience a force, as we know that force is a vector quantity hence it requires both magnitude and direction for it's complete description. The direction of this electric force that acts on a charge is given by the direction of the electric field in the space. In case the charge is negatively charged electric force will act on it in the direction opposite to the direction of electric field at the point.

A crate is resting on a ramp that is inclined at an angleabove
the horizontal. As the angle is incrased the crateremains in place
until the angle reaches a value of 38degrees. Then the crate begins
to slide down the slope. (a) Determine the coefficient of static
friction between the crateand the ramp surface. (b) The coefficient
of kinetic frictionbetween the crate and the ramps surface is
0.600. Find theacceleration of the moving crate.

Answers

Answer:

1.4 m/s^2

Explanation:

The angle of repose or the angle of sliding is defined as the angle of inclined plane with the horizontal at which an object placed on the plane is just start to slide down.

The relation between the angle of friction and the coefficient of static friction is given by

[tex]\mu _{s}=tan\theta[/tex]

[tex]\mu _{s}=tan38=0.78[/tex]

Thus, the coefficient of static friction is 0.78.

(b) As the crate is moving down, then the friction force is kinetic friction.

The force acting along the plane downward = mg Sinθ

The normal reaction, N = mg cosθ

The friction force acting upward along the plane, f = μk N = μk mg Cosθ

The net force acting along the plane downward

Fnet = mg Sinθ - μk mg Cosθ

According to the newton's second law, Fnet = mass x acceleration

so,  m x a = mg Sinθ - μk mg Cosθ

a = g Sinθ - μk g Cosθ

here, μk = 0.6 and θ = 38°, g = 9.8 m/s^2

By substituting the value, we get

a = 9.8 ( Sin38 - 0.6 x Cos 38)

a = 9.8 (0.6156 - 0.6 x 0.788)

a = 1.4 m/s^2

Final answer:

The coefficient of static friction between the crate and the ramp surface is 0.781. The acceleration of the moving crate is 5.88 m/s².

Explanation:

To find the coefficient of static friction, we need to calculate the tangent of the angle at which the crate begins to slide. The tangent of 38 degrees is approximately 0.781. Since the crate remains in place until this angle, the coefficient of static friction must be equal to or greater than this value. Therefore, the coefficient of static friction between the crate and the ramp surface is 0.781.

To find the acceleration of the moving crate, we can use the formula a = μk * g, where a is the acceleration, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity. Given that the coefficient of kinetic friction is 0.600, and the acceleration due to gravity is approximately 9.8 m/s², we can substitute these values into the formula to find the acceleration.

Substituting the values, we get a = (0.600) * (9.8) = 5.88 m/s². Therefore, the acceleration of the moving crate is 5.88 m/s².

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A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of the electrostatic force between those two parts?

Answers

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= [tex]\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}[/tex]

= 1.72 x 10³ N.

Consider a physical pendulum with length of 99.5 cm and mass of 185 g. If the pendulum was released from an angle less than 10°, then calculate the angular speed of the pendulum. (g = 9.80 m/s^2)

Answers

Answer:

The angular speed of the pendulum is 3.138 rad/s.

Explanation:

Given that,

Length = 99.5 cm

Mass = 185 g

We need to calculate the angular speed

Using formula of angular speed

[tex]\omega=\sqrt{\dfrac{g}{l}}[/tex]

Where, l = length

g = acceleration due to gravity

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{9.80}{99.5\times10^{-2}}}[/tex]

[tex]\omega=3.138\ rad/s[/tex]

Hence, The angular speed of the pendulum is 3.138 rad/s.

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a 36-N force is applied to the left side of the left cube causing all three cubes to accelerate to the right. If the cubes are each subject to a frictional force of 6.0 N, what is the magnitude of the force exerted on the right cube by the middle cube in this case

Answers

Answer:

Force exerted on the right cube by the middle cube:

F= 12.02N : in the positive direction of the x axis( +x)

Explanation:

We apply Newton's second law for forces in the direction of the x-axis.

∑Fx= m*a

∑Fx:  algebraic sum of forces ( + to the right, - to the left)

m: mass

a: acceleration

Forces (x) in total mass : Newton's second law

We apply Newton's first law for forces in the direction of the y-axis.

∑Fx= mt*a   , mt: total mass = 6*3= 18 kg

36-6= 18*a

[tex]a=\frac{36-6}{18} = 1.67 \frac{m}{s}[/tex]

Forces (y) in total mass : Newton's first law

∑Fy= 0

Nt-Wt=0   , Nt=Wt

Nt: total normal   ,  Wt= total Weight: mt*g  , g: acceleration due to gravity

Wt=18*9,8=176.4 N

Nt=176.4 N

Calculation of the coefficient of kinetic friction

μk=Ff/Nt Ff: friction force = 6 N

μk=6/176.4 = 0.034

Forces (x) on the first block (on the right): Newton's second law

∑Fx= m₁*a

F-Ff₁= 6*1.67 , Equation  (1)

F:Force exerted on the right cube by the middle cube

Ff₁= μk*N₁ , N₁=W₁ = 9.8*6= 58.8 N

Ff₁= 0.034*58.8 = 2 N

In the  Equation  (1):

F-2= 10.02

F= 2+10.02= 12.02N

F= 12.02N

An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following statements best describes the electric field at a distance r from the electron? The electric field is directed toward the electron and has a magnitude of ke/r. The electric field is directed away from the electron and has a magnitude of ke/2. The electric field is directed toward the electron and has a magnitude of ke/? The electric field is directed toward the electron and has a magnitude of ke?/r. The electric field is directed away from the electron and has a magnitude of ke/r.

Answers

Explanation:

The charge on the electron is, [tex]q=-1.6\times 10^{-19}\ C[/tex]

The electric field at a distance r from the electron is :

[tex]E=k\dfrac{q}{r^2}[/tex]

Where

k is the electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}[/tex]

We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.

So, the correct option is " the  electric field is directed toward the electron and has a magnitude of [tex]k\dfrac{q}{r^2}[/tex]. Hence, this is the required solution.

Two cars drive on a straight highway. At time t=0, car 1 passes mile marker 0 traveling due east with a speed of 20.0 m/s . At the same time, car 2 is 0.80 km east of mile marker 0 traveling at 26.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 0.40 m/s^2 , and car 2 is slowing down with an acceleration of magnitude 0.80 m/s^2 . a.) Write x-versus-t equations of motion for both cars, taking east as the positive direction.
b.) At what time do the cars pass next to one another?

Answers

Answer:

Explanation:

a ) Equation for car 1

X = 20 t + 1/2 x 0.40 t² ( initial velocity is 20 and acceleration is 0.4 )

Equation for car 2

X = 800 - ( 26 t - 1/2 x0.80 t² ) [ when t = 0 , X = 800 m and  with time t , X decreases ]

b ) Let after time t  they  meet , then X will be equal for both of them at t.

20 t + 1/2 x 0.40 t² = 800 - ( 26 t - 1/2 x0.80 t² )

0.20 t²- 46 t +800 = 0

t = 19 s and 211 s .

Final answer:

The equation of motion for car 1 is x = 20t + 0.2t^2. For car 2, it's x = 800 + 26t - 0.4t^2. To find when the cars meet each other, set these equal to each other and solve for t.

Explanation:

The x-versus-t equations of motion would be determined by rate of movement and acceleration. For car 1, which is accelerating positively, the equation of motion is x = v1*t + 0.5*a1*t^2 (where v1 = initial velocity of car 1, a1 = acceleration of car 1, t is time). So, it would be x = 20t + 0.5 * 0.4 * t^2 = 20t + 0.2t^2.

For car 2, decelerating negatively and starting at 0.80 km east of the origin (which equates to 800m), the equation is x = x2 + v2*t - 0.5*a2*t^2. This works out to x = 800 + 26t - 0.5 * 0.8 * t^2 = 800 + 26t - 0.4t^2.

To find the time (t) when cars pass each other, we equate the two displacements: 20t + 0.2t^2 = 800 + 26t - 0.4t^2. Solving that equation gives us the time in seconds when the cars will meet next to each other.

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A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g of ethyl alcohol at 20 degrees. What is the final temperature?

Answers

Final answer:

The final temperature can be calculated using the principle of conservation of energy. Using the formulas Q = mcΔT and Q = mcΔT, we can find the heat lost by the iron and the heat gained by the ethyl alcohol. The final temperature of all 3 substances are, T_f = 12.08 °C.

Explanation:

To calculate the final temperature of the substances, we can use the principle of conservation of energy. The heat gained by one substance is equal to the heat lost by another substance. First, we calculate the heat lost by the iron using the formula Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Then, we calculate the heat gained by the ethyl alcohol using the same formula. Finally, we use the principle of conservation of energy to find the final temperature.

We have:

Iron: m = 200 g, c = 452 J/g °C, Ti = 120 °CCopper: m = 150 g, c = 387 J/g °C, Ti = -50 °CEthyl alcohol: m = 300 g, c = 2.30 J/g °C, Ti = 20 °C

Using the formula Q = mcΔT, we can calculate:

Heat lost by iron: Qiron = (200 g)(452 J/g °C)(120 °C - T_f)Heat gained by ethyl alcohol: Qethyl alcohol = (300 g)(2.30 J/g °C)(T_f - 20 °C)

Since the heat lost by the iron is equal to the heat gained by the ethyl alcohol, we can set up the equation:

(200 g)(452 J/g °C)(120 °C - T_f) = (300 g)(2.30 J/g °C)(T_f - 20 °C)

To solve for T_f, we can simplify and rearrange the equation:

(90400 - 452T_f) J = (T_f- 13800) J

Combining like terms:

13800 J = 1142T_f J

Dividing both sides by 1142:

T_f = 12.08 °C

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A container is filled to the brim with 1.5 L of mercury at 20°C. As the temperature of the container and mercury is increased to 55°C, a total of 7.8 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container.

Answers

Answer:

The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]

Explanation:

Given that,

Volume = 1.5 l

Temperature = 20°C

Increased temperature = 55°C

Volume = 7.8 mL

We need to calculate the linear expansion coefficient of the material

Using formula of  linear expansion

[tex]\Delta V=V_{0}\beta \Delta T[/tex]

[tex]\Delta V=V_{0}(3\alpha \Delta T)[/tex]

[tex]\alpha =\dfrac{\Delta V}{3V_{0}\Delta T}[/tex]

Put the value into the formula

[tex]\alpha=\dfrac{7.8\times10^{-3}}{3\times1.5\times(55-20)}[/tex]

[tex]\alpha=0.00004952[/tex]

[tex]\alpha=4.952\times10^{-5}\ K^{-1}[/tex]

Hence, The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]

A traveler covers a distance of 217 miles in a time of 8 hours 32 minutes. (a) What is the average speed for this trip?

Answers

Answer:

Average speed, v = 11.37 m/s

Explanation:

Given that,

The distance covered by the traveler, d = 217 miles = 349228 meters

Time taken, [tex]t = 8\ hours \ 32\ minutes =8\ h+\dfrac{32}{60}\ h=8.53\ h[/tex]

or t = 30708 s

We need to find the average speed for this trip. The average speed is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{349228\ m}{30708\ s}[/tex]  

v = 11.37 m/s

So, the average speed for this trip is 11.37 m/s. Hence, this is the required solution.

What additional information is obtained from measurements on an ordinal scale compared to measurements on a nominal scale? Comparison of the magnitudes of two measurements (such as one being twice as large as the other) Direction of difference (greater or less) between two measurements Magnitude of difference between two measurements

Answers

No idea, if you go on this website called Brainly they have all of the answers for you.

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vector at the point P, a distance d above the midpoint of the wire. The field E-vector points along one of the primary axes, yWhat is the magnitude E of the electric field at point P? Throughout this part, express your answers in terms of the constant k, defined by k=1/(4πε)

Answers

Answer:

[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

[tex]E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta[/tex]

[tex]sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}[/tex]

[tex]r^2=d^2+x^2[/tex]

So

[tex]E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}[/tex]

Now by integrating above equation

[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it starts to careen down the steeply sloped track. In one event, the sled reaches a top speed of 9.2 m/s before starting down the initial part of the track, which is sloped downward at an angle of 4.0°. What is the sled's speed after it has traveled the first 140 m?

Answers

Answer:

[tex]v_f = 16.6 m/s[/tex]

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

[tex]F_g = mg sin\theta[/tex]

so the acceleration due to gravity along the plane is given as

[tex]a = \frac{F_g}{m}[/tex]

now we have

[tex]a = g sin\theta[/tex]

[tex]a = (9.81 sin4.0)[/tex]

[tex]a = 0.68 m/s^2[/tex]

now we know that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 9.2^2 = 2(0.68)(140)[/tex]

[tex]v_f = 16.6 m/s[/tex]

Velocity of a object is the ratio of distance traveled by the object with the time taken

The sled's speed after it has traveled the first 140 m is 16.6 m/s.

What is velocity of a object?

Velocity of a object is the ratio of distance traveled by the object with the time taken.

Given information-

The top speed reached by the sled is 9.2 m/s.

The angle of the slope is 4 degrees downward.

Distance traveled by the sled is 140 meters.

The force acting on a body is the product of mass and its acceleration.

The acceleration of the inclined plane with this definition can be given as,

[tex]a=\dfrac{F_g}{m} =\dfrac{mg\sin\theta}{m} \\a=g\sin\theta[/tex]

Here, [tex]F_g[/tex] is the force due to gravity and [tex]a[/tex] is the acceleration of the body.

Put the values in the above equation as,

[tex]a=9.81\times \sin(4^o)\\a=0.68 \rm m/s^2[/tex]

Thus the acceleration of the body is 0.68 m/s squared.

Now the sled's speed can be find out using the equation of motion as,

[tex]2ad=v_f^2-v_i^2[/tex]

As the initial speed is 9.2 seconds. Thus,

[tex]2\times0.68\times140=v_f^2-9.2^2\\v_f=16.6 \rm m/s[/tex]

Hence the sled's speed after it has traveled the first 140 m is 16.6 m/s.

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A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067 g d) 0.015 g

Answers

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

[tex]P = mv[/tex]

[tex]m = \dfrac{P}{v}[/tex]

Where, P = momentum

v = speed

Put the value into the formula

[tex]m = \dfrac{2.8}{187}[/tex]

[tex]m = 0.015\ kg[/tex]

[tex]m = 15\ g[/tex]

Hence, The mass of the bullet is 15 g.

Is a sonic boom heard by the passengers traveling in a supersonic aircraft? Why or why not? a.) they do not experience any shift in frequency because they are traveling at the same speed as that of the sound source.
b.) They do not experience any shift because they are traveling at a lower speed than that of the sound source.
c.) They will experience a small shift in frequency because they are traveling at a lower speed than that of the sound source.
d.) They will experience a large shift in frequency because they are traveling at a greater speed than that of the sound source.

Answers

Answer:

No the sonic boom is not heard by the passenger travelling in a supersonic aircraft. They do not experience any shift in the frequency as they are travelling at the speed of the source itself.Thus option 'a' is correct answer.

Explanation:

Sonic boom is a shock wave of sound associated when a source travels faster than the speed of sound in the ambient conditions. In a shock wave associated with the sonic boom there exists high gradients of pressure, density and temperature of air across the shock wave. The shock wave travels at the speed of the sound but the source (in our case the passenger) travels at a faster speed than the sound in ambient conditions.

The average velocity for a trip has a positive value. Isit
possible for the instantaneous velocity at any point during thetrip
to have a negative value? Justify the answer.

Answers

Answer:

Yes, it's possible.

Explanation:

The average velocity is a mean value:

[tex]Vavg=\frac{displacement}{time taken}[/tex].

during that displacement, it may occur that the acceleration would negative at any time so at that moment if the velocity goes in the same direction with the acceleration, the velocity will be negative, it may take just a few moments and then go positive again. The velocity can also take negative values if for a moment the object was going backward (opposite direction). so the average velocity only means that the major of the velocity was positive.

A golf ball is hit into the air, but NOT Staight Up, and encounters no significant air resistance. Which statement accurately describes its motion while it is in air? On the way up, both its horizontal and vertical velocity components are decreasing, on the way down they are both increasing. Its acceleration is zero at the highest point. Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change. On the way up its acceleration is 9.8m/s2 upward, on the way down its acceleration is 9.8m/s2 downward, Its velocity is zero at the highest point.

Answers

Answer:

Answered

Explanation:

A) False , horizontal velocity component is constant since there's no horizontal acceleration component

B) False, acceleration remains constant through out and only verticle acceleration is there horizontal acceleration is zero.

C) True , Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change ( because of acceleration in verticle direction)

D)False, the acceleration is constant through the motion.

E)False, Vertical velocity is Zero but there will be Horizontal velocity.

Two point charges are located on the x-axis. One has a charge of 1.77 μC and is located at x = 0.0 m, and the other has a charge of -4.09 μC and is located at x = 15.1 m. At what location on the x-axis (other than at infinity) would the electric force on a third point charge of 3.32 μC be zero?

Answers

Answer:

x = -29.032m

Explanation:

Since the third charge is positive, it cannot be between the other two charges, because it would be repelled by the positive one and attracted by the negative one, so the electric force would never be zero.

This leaves only two options: To the left of the positive one or to the right of the negative one.

If it was located on the right of the negative charge, the force of the positive charge would be weaker because of both the distance is larger and its charge is smaller than the negative charge. So, there is no point the would make the result force equal zero.

This means that the third charge has to be at the left of the positive charge. With this in mind, we make the calculations:

[tex]F_{13}=K*\frac{Q_{1}*Q_{3}}{d^{2}} =F_{23}=K*\frac{Q_{2}*Q_{3}}{(x_{2}+d)^{2}}[/tex]

Replacing the values of Q1=1.77, Q2=4.09, X2=15.1, we solve for d and get two possible results:

d1 = 29.032m   and d2 = -5.99m

Since we assumed in our formula that the third charge was on the left of the positive charge, the distance d has to be positive so that our final result can be a negative position. This is X = -d

This way, we get:

X = -29.032m

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, your speed is decreased by the water's speed. The current in a river flows at 0.52 m/s. In still water you can swim at 1.73 m/s. If you swim downstream a certain distance, then back again upstream, how much longer, in percent, does it take compared to the same trip in still water?

Answers

Swimming downstream and upstream in a river with a current of 0.52 m/s results in approximately an 86% increase in time compared to swimming the same distance in still water at a speed of 1.73 m/s.

1. Understanding the problem:

You can swim 1.73 m/s in still water.

The river current flows at 0.52 m/s.

We need to find the percentage increase in time to swim downstream and back upstream compared to swimming the same distance in still water.

2. Effective speeds:

Downstream: Your speed increases with the current.

Effective speed = swimming speed + current speed = 1.73 m/s + 0.52 m/s = 2.25 m/s

Upstream: Your speed decreases against the current.

Effective speed = swimming speed - current speed = 1.73 m/s - 0.52 m/s = 1.21 m/s

3. Calculating time for each leg:

Let D be the distance for the entire round trip.

Time downstream: D / 2.25 m/s

Time upstream: D / 1.21 m/s

4. Finding the time difference:

Time difference = Time upstream - Time downstream

Time difference = (D / 1.21 m/s) - (D / 2.25 m/s)

5. Calculating percentage increase:

To find the percentage increase in time compared to still water, divide the time difference by the time taken to swim downstream and multiply by 100.

Percentage increase = [(D / 1.21 m/s) - (D / 2.25 m/s)] / (D / 2.25 m/s) * 100

6. Simplifying the equation:

After simplifying, you'll find the percentage increase to be approximately 86%.

Therefore, it takes approximately 86% longer to complete the round trip compared to swimming the same distance in still water.

A cyclist reaches the bottom of a hill with a speed of 18 m/s. Neglecting air resistance and other friction, to what maximum height can they coast up the hill without pedalling? a) 19 m b) 20 m c) 17 m d) 18 m

Answers

Answer:

c) 17 m

Explanation:

From the conservation of energy.

Change in kinetic energy= change in potential energy

Now,

[tex]\Delta KE=\Delta PE[/tex]

Now according to the situation.

[tex]\frac{1}{2}mv^{2}=mgh\\h=\frac{v^{2} }{2g}[/tex]

Here, v is the velocity, m is the mass of an object, h is the height, g is the acceleration due to gravity.

Given that the speed of the bicycle is

[tex]v=18m/s[/tex]

And the acceleration due to gravity is

[tex]g=9.8 m/s^{2}[/tex]

Substitute these values.

[tex]h=\frac{18^{2} }{2\times 9.8}\\h=16.53m[/tex]

Which means it is equivalent to 17 m

Therefore, the maximum height is 17 m

Explain the meaning of the term "potential divider" as applied to the circuit?

Answers

Answer and Explanation:

A potential divider, as the name itself gives  a clear description of a simplified circuit which divides the potential across different circuit elements.

The potential divider circuit divides the potential depending upon the respective value of the elements of the circuit,

Given circuit diagram shows a potential divider circuit.

The working of this circuit is as :

[tex]V' = \frac{R'}{R + R'}V[/tex]

where

V = Input voltage

V' = output voltage

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 7.5 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.

Answers

Answer:

450 pm

Explanation:

The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:

a = v^2 / r

The equation for the electric force is:

F = q1 * q2 / (4 * π * e0 * r^2)

Where

q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C

e0: electric constant (8.85*10^-12 F/m)

If we divide this force by the mass of the electron we get the acceleration

me = 9.1*10^-31 kg

a = q1 * q2 / (4 * π * e0 * me * r^2)

v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)

We can simplify r

v^2 = q1 * q2 / (4 * π * e0 * me * r)

Rearranging:

r = q1 * q2 / (4 * π * e0 * me * v^2)

r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answers

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

[tex]h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2[/tex]    .... (2)

By equation the equation (1) and (2), we get

[tex]41.7=1.12 u +4.9 \times 1.12^{2}[/tex]

u = 31.75 m/s

A golfer takes three strokes to putt a golf ball into a hole. On the first stroke, the ball moves 4.8 m due east. On the second, it moves 2.7 m at an angle 20° north of east. On the third, it moves 0.50 m due north. If the golfer had instead hit the ball directly into the hole on the first stroke, what would have been the magnitude (in m) and direction of the ball's displacement? Give the direction as a positive angle measured counterclockwise from due east.

Answers

The magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.

1. First stroke:

 [tex]Displacement \( \vec{d}_1 = 4.8 \, \text{m} \) due east.[/tex]

2. Second stroke:

[tex]Eastward component: \( 2.7 \, \text{m} \times \cos(20^\circ) \approx 2.564 \, \text{m} \) due east.\\ Northward component: \( 2.7 \, \text{m} \times \sin(20^\circ) \approx 0.924 \, \text{m} \) due north.\\\\ Displacement \( \vec{d}_2 = 2.564 \, \text{m} \) due east and \( 0.924 \, \text{m} \) due north.[/tex]

3. Third stroke:

  Displacement[tex]\( \vec{d}_3 = 0.50 \, \text{m} \)[/tex] due north.

Now let's calculate the total eastward and northward displacements:

Total eastward displacement:[tex]\( 4.8 \, \text{m} + 2.564 \, \text{m} = 7.364 \, \text{m} \)[/tex]

Total northward displacement: [tex]\( 0.924 \, \text{m} + 0.50 \, \text{m} = 1.424 \, \text{m} \)[/tex]

Using the Pythagorean theorem, the magnitude of the total displacement is:

[tex]Magnitude = \( \sqrt{(\text{Total eastward displacement})^2 + (\text{Total northward displacement})^2} \)Magnitude = \( \sqrt{7.364^2 + 1.424^2} \approx 7.579 \, \text{m} \)[/tex]

The direction of the total displacement is given by:

[tex]Direction = \( \tan^{-1}\left(\frac{\text{Total northward displacement}}{\text{Total eastward displacement}}\right) \)\\Direction = \( \tan^{-1}\left(\frac{1.424}{7.364}\right) \approx 11.08^\circ \) counterclockwise from due east.[/tex]

So, if the golfer had hit the ball directly into the hole on the first stroke, the magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.

Learn more about displacement, here:

https://brainly.com/question/11934397

#SPJ12

Final answer:

The golfer's ball would have a direct displacement of approximately 7.434 meters in a direction of roughly 11° north of east if it were hit directly into the hole.

Explanation:

To determine the direct displacement and direction of the ball if the golfer had hit it directly into the hole, we need to sum up the three given vectors. The first is 4.8 m due east, the second is 2.7 m at a 20° angle north of east, and the third is 0.50 m due north.

First, decompose the second stroke into its north and east components. The east component is 2.7 cos(20°) m and the north component is 2.7 sin(20°) m.

East component of the second stroke: 2.7 cos(20°) ≈ 2.536 m

North component of the second stroke: 2.7 sin(20°) ≈ 0.923 m

Now, sum the components in each direction:

Total east displacement: 4.8 m + 2.536 m = 7.336 m

Total north displacement: 0.923 m + 0.50 m = 1.423 m

Finally, calculate the magnitude and direction of the resultant displacement vector:

Magnitude (√(x² + y²)): √(7.336² + 1.423²) ≈ 7.434 m

Direction (tan⁻¹(y/x)): tan⁻¹(1.423/7.336) ≈ 11° north of east

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