A physics student stands on a bathroom scale in an elevator which is not moving. The scale reads 604 N. The elevator then accelerates downward at 1.98 m/s2. What does the scale read during this acceleration?

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]

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

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:

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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Answer:

a) a = 2000 g ,  b)  y = 4 cm

Explanation:

a) We can solve this part using the free fall relations

Let's start by finding the speed with which it reaches the ground, just before it starts to decelerate

           v₁² = v₀² + 2 g y

as it comes out of rest the initial velocity is v₀ = 0

           v₁ = √ 2g y

           v₁ = √ (2 9.8 2)

           v₁ = 6.26 m / s

in the braking part the distance is y₂ = 1 mm = 0.001 m

          v² = v₁² - 2 a y₂

       when stopping its velocity is zero v = 0

        a = v₁² / 2y₂

        a = 6.26² / (2 0.001)

        a = 1.96 10⁴ m / s²

This is the braking acceleration.

Let's look for its relationship with the acceleration of gravity

          a / g = 1.96 10⁴ / 9.8

          a / g = 2000

          a = 2000 g

b) to reduce the injury to less than 25%, the maximum acceleration that must be a = 50g = 490 m / s²

Let's find the distance it must travel to have this acceleration

          v² = v₁² - 2 a y

          y = (0 + v₁²) / 2a

          y = 6.26² / (2 490)

          y = 0.04 m

          y = 4 cm

therefore we must have a helmet that deforms 4 cm before the head stops

Most bicycle helmets provide this protection since they are quite elongated

c) The main error in the calculations is not taking into account the  

resistance with the air, in this sense the speed would be less than the calculated one and our calculations would be over estimated.

The second factor is that in a fall there may be an initial velocity different from zero, so the velocity would be higher and the acceleration as well, in this case our calculations are underestimated.

Answer:

b

Explanation:

Answer:

false

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)

Answer:

the order from highest to lowest is      C> B> F> E> A> D

Explanation:

This is an exercise in calorimetry where the heat given off by the copper cylinder is equal to the heat absorbed by the water  

     Q_c = m_Cu ce_Cu ΔT_Cu  

    Q_a = m_water Ce_water ΔT  

    Q_c = Q_a  

    m_Cu ce_Cu (T_o-T_f) = m_water ce_water (T_f - T_i)

we clear the final temperature and substitute the values

With this expression we can know the final temperature of the system, let's substitute the values ​​that are constant throughout the calculation

With this expression we can know the final temperature of the system, let's substitute the values ​​that are constant throughout the calculation

a) m = 0.250kg, To = 30ºC  

      T_{fa}=(387 0.250 30 + 376740) / (15697.5 + 387 0.250)  

      T_{fa}= 379642.5 / 15794.25  

      T_{f} = 24.04ºC  

b) m = 0.500 kg, To = 60ºC  

     T_{fb} = (387 0.500 60 + 376740) / (15697.5 + 387 0.500)

     T_{fb} = 388350/15891  

     T_{fb = 24.44 ° C

c) m = 0.750 kg, To = 90ºC  

    T_{fc}= (387 0.750 90 + 376740) / (15697.5 + 387 0.750)  

    T_{fc}= 402862.5 / 15987.75  

    T_{fc} = 25.20ºC  

d) m = 0.500 kg, To = 15ºC  

    T_{fd} = (387 0.500 15 + 376740) / (15679.5 + 387 0.500)

    T_{fd} = 379642.5 / 15891  

    T_{fd} = 23.89ºC

the order from highest to lowest is      C> B> F> E> A> D

Answer:

The correct option is a: 214 nm.

Explanation:

The wavelength (λ) of the light can be calculated as follows:

[tex] \lambda = \frac{yd}{nD} [/tex]   (1)

Where:

y: is the distance of the 7th bright fringe on the screen from the central fringe = 10.0 cm

d: is the distance between the slits = 3.00x10⁻⁵ m

D: is the distance between the screen and the slits = 2.00 m

n = 7

By entering all the above values into equation (1) we have:

[tex] \lambda = \frac{yd}{nD} = \frac{0.10 m*3.00 \cdot 10^{-5} m}{7*2.00 m} = 214.3 nm [/tex]

Therefore, the correct option is a: 214 nm.

I hope it helps you!

Explanation:

F=GMm/d^2

Distance between friends and us <<<<<Distance between centre of earth and us

Answer:

kinsey sexuailty doesn't matter because that is his or her  

Explanation:

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:

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:

b. increases above 750N

Explanation:

Okay

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]

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:

Transformers do not allow DC input to flow through. This is known as DC isolation. This is because a change in current cannot be generated by DC

Explanation:

Hope this helps

Answer:

a change in current cannot be generated by DC;

Answer:

B

Explanation:

Answer:

The current will flow counter-clockwise and be decreasing

Explanation:

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

Answer:

Try B or C if I'm wrong sorry

Explanation:

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.

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.

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:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60  hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

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%

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 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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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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