A bird uses 10 N of force to pull a worm out of the ground a distance of 3 inches. How much work did the bird do?

Answer:

120 mph

Explanation:

Given:

Δx = 0.25 mi

v₀ = 0 mi/s

t = 15 s

Find: v

Δx = ½ (v + v₀) t

0.25 mi = ½ (v + 0 mi/s) (15 s)

v = 0.0333 mi/s

v = 120 mi/h

The normal force applied by the wall in the ladder will be directed to the left. Hence, option (B) is correct.

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

In this context, normal refers to perpendicular in a geometric sense rather than "usual" or "expected" as it does in everyday parlance. Gravity acts on a person standing stationary on a platform; gravity would otherwise draw the person down into the Earth's core absent a countervailing force from the resistance of the platform's molecules, known as the "normal force."

As normal force acts perpendicular to the surface, the normal force applied by the wall in the ladder will be directed to the left.

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Answer:

F = 17270 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by the acceleration.

ΣF = m*a

where:

F = force [N]

m = mass = 3454 [kg]

a = acceleration = 5[m/s^2]

Now replacing:

F = 3454*5

F = 17270 [N]

Answer:

LT 23 don't ask how I got it

Answer:

The required work done is [tex]6.5\times10^{9}\ J[/tex]

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

[tex]U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}[/tex]

Put the value into the formula

[tex]U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}[/tex]

[tex]U_{A}=-3.44\times10^{10}\ J[/tex]

We need to calculate the potential energy

Using formula of potential

[tex]U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}[/tex]

Put the value into the formula

[tex]U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}[/tex]

[tex]U_{B}=-2.14\times10^{10}\ J[/tex]

We need to calculate the value of [tex]k_{A}[/tex]

Using formula of [tex]k_{A}[/tex]

[tex]k_{A}=-\dfrac{1}{2}U_{A}[/tex]

Put the value into the formula

[tex]k_{A}=\dfrac{1}{2}\times3.44\times10^{10}[/tex]

[tex]k_{A}=1.72\times10^{10}\ J[/tex]

We need to calculate the value of [tex]k_{B}[/tex]

Using formula of [tex]k_{B}[/tex]

[tex]k_{B}=-\dfrac{1}{2}U_{B}[/tex]

Put the value into the formula

[tex]k_{B}=\dfrac{1}{2}\times2.14\times10^{10}[/tex]

[tex]k_{B}=1.07\times10^{10}\ J[/tex]

We need to calculate the work done

Using formula of work done

[tex]W=\Delta K+\Delta U[/tex]

[tex]W=(k_{B}-k_{A})+(U_{B}-U_{A})[/tex]

[tex]W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})[/tex]

[tex]W=\dfrac{1}{2}(U_{B}-U_{A})[/tex]

Put the value into the formula

[tex]W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})[/tex]

[tex]W=6.5\times10^{9}\ J[/tex]

Hence, The required work done is [tex]6.5\times10^{9}\ J[/tex]

Answer:

B

Explanation:

The rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

Given the data in the question;

Rotational or kinetic energy; [tex]E_{rotational} = \ ?[/tex]

Rotational energy or angular kinetic energy is simply kinetic energy due to the rotation of a rigid body.

It is expressed as;

[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]

Where [tex]I[/tex] is the moment of inertia around the axis of rotation and [tex]w[/tex] is the angular speed or velocity.

For the moment of inertia around the axis of rotation.

[tex]I = \frac{1}{2}mr^2[/tex]

Hence

[tex]E_{rotational} = \frac{1}{2}(\frac{1}{2}mr^2)w^2 \\\\E_{rotational} = (\frac{1}{4}mr^2)w^2[/tex]

Now, we substitute our given values into the above equation to find the rotational or kinetic energy.

[tex]E_{rotational} = (\frac{1}{4}*3.0kg * (0.1m)^2) * (3.5rad/s)^2 \\\\E_{rotational} = 0.0075kgm^2 * 12.25rad/s^2\\\\E_{rotational} = 0.092kg.m^2/s^2\\\\E_{rotational} = 0.092J[/tex]

Therefore, the rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

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Answer:

b

Explanation:

Answer:

false

Explanation:

Answer:

S = Vo t + 1/2 a t^2 = 1/2 a t^2       V0 = 0 for a standing start

t = (2 S / a)^1.2

t = (2 * 30 / (2.0) =  (30)^1/2 = 5.5 sec

The velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

The velocity of the cylinder can be determined by applying the principle of conservation of energy as shown below;

Ei = Ef

¹/₂k(xi)² + mgh = ¹/₂mv² + ¹/₂k(xt)²

where;

¹/₂k(xi)² = -mgh + ¹/₂mv² + ¹/₂k(xt)²

¹/₂(1050)(0.076)² = -(45.35)(9.8)(0.0127) + ¹/₂(45.35)v² + ¹/₂(1050)(0.165)²

3.032 = -5.64 + 22.68v² + 14.29

3.032 = 8.65 + 22.68v²

-5.62 = 22.68v²

|v| = 0.497 m/s

Thus, the velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

The complete question is below:

The spring has a stiffness of 6 lb/in and the mass of the load is 100 lb.

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The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

Answer:

Try B or C if I'm wrong sorry

Explanation:

Hi there!

Hi there!

We can begin by simplifying the work-energy theorem for Crate 2.

Since there is no friction, there is no energy dissipated. Thus, the initial energy is equal to the final energy.

Initially, we only have gravitational potential energy (U = mgh), and when the box has fully slid down, it only has kinetic energy (KE = 1/2mv²), therefore:
[tex]E_i = E_f\\\\mgh = \frac{1}{2}mv^2[/tex]

We can cancel out the mass and solve for velocity.

[tex]gh = \frac{1}{2}v^2\\\\v^2 = 2gh \\\\v = \sqrt{2gh}[/tex]

We must use right triangle trigonometry to solve for the HEIGHT given the ramp's length (hypotenuse).

We can use sine:
[tex]sin\theta = \frac{\text{h}}{L} \\\\Lsin\theta = h = 6 * sin(25) = 2.5357 m[/tex]

Now, solve for velocity.

[tex]v = \sqrt{2(9.8)(2.5357)} = 7.05 \frac{m}{s}[/tex]

Since this is 2.5 times the speed of the first crate, we know that the final velocity of crate 1 is:

[tex]v_1 = \frac{v}{2.5} = 2.82 \frac{m}{s}[/tex]

Crate 1:
In this instance, we have friction. Recall the following.

[tex]F_f = \mu N[/tex]

On an incline, the normal force is equivalent to the cosine of the force of gravity, so:
[tex]N = mgcos\theta[/tex]

Now, create an equation for the force due to friction.

[tex]F_f = \mu mgcos\theta[/tex]

The work done by any force is:
[tex]W = F \cdot d\\\\W_f = \mu mgdcos\theta[/tex]

In this instance, d = the ramp's length, or 6 m.

Now, we can use the work-energy theorem.

Ei = Ef

However, there is energy dissipated; we can call this Wf (Work due to friction). Therefore:
Ei - Wf = Ef

Now, we can rearrange to solve for Wf:
Ei - Ef = Wf

Like above, there is initially only GPE (U = mgh) and finally only KE (K = 1/2mv²), so:
[tex]mgh - \frac{1}{2}mv^2 = \mu mgdcos\theta[/tex]

Solve for the coefficient of friction. Begin by canceling out the mass and multiplying all terms by 2:
[tex]2mgh - mv^2 = 2\mu mgdcos\theta\\\\2gh - v^2 = 2\mu gdcos\theta\\\\\mu = \frac{2gh - v^2}{2gdcos\theta}\\\\\mu = \frac{2(9.8)( 2.5357)- (2.82)^2}{2(9.8)(6)cos(25)}[/tex]

Evaluate:
[tex]\boxed{\mu = 0.39}[/tex]

Hi there!

Recall the equation for the voltage of a discharging capacitor:

[tex]V_C(t) = V_0e^{-\frac{t}{\tau}}[/tex]

V₀ = Initial voltage of the capacitor (V)
t = Time (s)
τ or RC = Time Constant (s)

With the given information, we can plug in the value for V₀:
[tex]V_C(t) = 3.2e^{-\frac{t}{\tau}}[/tex]

We are given that at t = 20 ms (0.02 s), the voltage of the capacitor is 0.8V. We can use this to solve for the time constant (τ).

[tex]0.8 = 3.2e^{-\frac{0.02}{\tau}}\\\\0.25 = e^{-\frac{0.02}{\tau}}[/tex]

Take the natural log of both sides and solve.

[tex]ln(0.25) = -\frac{0.02}{\tau}\\\\-1.3863 = -\frac{0.02}{\tau}\\\\\tau = \frac{0.02}{1.3863} = 0.0144 s[/tex]

Now, we can use this time constant to solve for the time taken for the voltage to drop from 0.8 V to 0.2 V. Solve for the time taken for the capacitor's voltage to drop to 0.2 V:

[tex]0.2= 3.2e^{-\frac{t}{0.0144}}\\\\0.0625 = e^{-\frac{t}{0.0144}}\\\\ln(0.0625) = -\frac{t}{0.0144}\\\\t = (-2.773)(-0.0144) = 0.04 s[/tex]

Now, subtract the times:
[tex]0.04 - 0.02 = 0.02 = \boxed{20 ms}[/tex]

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

vi=0, vf=80km/hr=22m/s, m=1.5x107kg, F = 7.5x105N.

a=F/m, t=(vf-vi)/a= (vf-vi)m/F = (22m/s)(1.5x107kg)/7.5x105N = 440 kg m/sN =  440 s or 7mins

Answer:

L = L0 (1 - v^2/c^2)     where L0 is proper length and L the measured length

L = 1.2 (1 - .87^2)^1/2 m = .59 m

Answer:

The moving object transfers kinetic energy with the object that isn't moving.

Explanation:

Kinetic energy is a form of energy that a object  or particle has by the reason of motion. Potential energy is a form of energy that a object or particle has by the reason of no movement. The object that is moving knocks down or collides with an object that isn't  moving, and the object that is moving transfers kinetic energy to potential energy, and the object that wasn't moving transfers potential energy to kinetic energy.

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

The maximum distance through which the fish falls is 0.25m.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

Given parameters:

Spring constant: k = 200 N/m.

Mass of the fish: m =2.6 kg.

Let the maximum distance through which the fish falls, i.e. the maximum stretched distance of the spring = x.

From law of conservation of energy:

Change in gravity potential energy  = change in potential energy

mgx = 1/2kx^2

2.6×9.8×h = 1/2*200*h^2

h(100h - 2.6×9.8) = 0

h = 0 or h = 2.6×9.8/100 = 0.25m

As h = 0 is the equilibrium position; the maximum distance through which the fish falls is 0.25m.

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The instantaneous power output at t = 3.7 s right answer is 99 W.

Power is defined by the amount of energy transferred over a period of time. Instantaneous power, on the other hand, refers to the power consumed at a point in time. Instantaneous power is an important metric in electronics. Instantaneous power is the power measured at a specific point in time.

The basic difference between average effort and instantaneous effort is that average effort is the ratio of total work time to total time. Although the instantaneous power is the limit of the average power. The statement of work does not state that the same amount of work he will complete in a second or an hour. Instantaneous power can be positive or negative. Positive instantaneous power means that energy is flowing from the source to the load, and negative instantaneous power means that energy is flowing from the load to the source.

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Answer:

When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake

Explanation:

I'm not entirely sure this is what you were looking for but I hope this helped!

PLEASE MARK ME AS BRAINLIEST

Explanation:

please mark me as brainlest

1.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

2.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

3.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

4.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

5.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

6.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

The orbital radius of the satellite orbiting planet Earth at the given linear speed is 26,000 km.

The orbital radius of the planet is the distance of the planet from the center of the Earth. The orbital radius of the planet is calculated as follows;

[tex]v = \sqrt{\frac{GM}{r} }[/tex]

where;

r = GM/v²

r = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴)/(3,914²)

r = 2.6 x 10⁷ m

r = 26,000 km

Thus, the orbital radius of the planet is 26,000 km.

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Answer:

v = ω R

number of seconds in 24 hrs = 24 * 3600 = 86400 s / da = T (period)

f = 1 / T

ω  = 2 Π f = 2 Π / T = 2 Π / 86400

v = 2 Π / 8.64E4 * 6.0E6 = 436 m / s

Check:

S = 2 Π * 6.0E6 = 3.77E7 m/da

S / T = 3.77E7 m/da / 8.64E4 s/da = 436 m / s

Answer:

1. Acceleration begins to slow down quickly

2. Kenetic

3. Mass

Answer:

Acceleration slows down

Kinetic

Mass

Explanation:

Answer:

bouncing  a baskletball

Explanation:

Answer:someone kicking a soccer ball.

Explanation:

Because the ball isn’t in motion until acted on by another object (the foot)