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qaws [65]
3 years ago
7

The kinetic energy of the 2 kg object when d equals 20 m is the same as when d is most nearly

Physics
1 answer:
EastWind [94]3 years ago
8 0

Explanation:

For the given question, distance is given on x-axis and force is given on y-axis. Both distance and force are showing an interval of 5 with starting point 0.

Hence, work in 5 m to 12.5 m is equal and opposite to 12.5 m to 20 m. Since, work is equal then it means that kinetic energy at 5 m will also be equal to the kinetic energy at 20 m.

Therefore, we can conclude that the kinetic energy of the 2 kg object when d equals 20 m is the same as when d is most nearly  5 m.

You might be interested in
Part 1 - Basic Equations
bearhunter [10]

Answer:

1. λ = 2 L, 2.  v = 2L f₁ , 3.    v = √ T /μ², 4.   μ = 2,287 10⁻³ kg / m , 5.   Δv / v = 0.058 , 6.    Δμ /  μ = 0.12 , 7. Δ μ = 0.3  10⁻³ kg / m ,

8.  μ = (2.3 ±0.3)  10⁻³ kg / m

Explanation:

The speed of a wave is

            v = λ f                1

Where f is the frequency and λ the wavelength

     

The speed is given by the physical quantities of the system with the expression

            v = √ T /μ²                   2

1) The fundamental frequency of a string is when at the ends we have nodes and a maximum in the center, therefore this is

                 L = λ / 2

                 λ = 2 L

2) For this we substitute in equation 1

              v = 2L f₁

3) let's clear from equation 2

             

The speed of a wave is

            v = λ f₁

Where f is the frequency and Lam the wavelength

The speed is given by the physical quantities of the system with the expression

           v = √ T /μ²                            2

4) linear density is

           μ = T / (2 L f₁)²

           μ = 5.08 / (2 0.812 29.02)²

           μ = 2,287 10⁻³ kg / m

We maintain three significant length figures, so the result is reduced to

           μ = 2.29 10⁻³ kg / m

5) the speed of the wave is

            v = 2 L f₁

The fractional uncertainty is

         Δv / v = ΔL / L + Δf₁ / F₁

         Δv / v = 0.02 / 0.812 + 1 / 29.02

         Δv / v = 0.024 + 0.034

         Δv / v = 0.058

6) the equation for linear density is

              μ = T / (2 L f₁)²

             Δμ / μ = 2 ΔL / L + 2Δf₁ / f₁

The tension is an exact value therefore its uncertainty is zero ΔT = 0

            Δμ / μ = 2 0.02 / 0.812 + 2 1 / 29.02

             Δμ /  μ = 0.12

7) absolute uncertainty

           Δ μ = e_{r}   μ

           Δ μ = 0.12 2.29 10⁻³ kg / m

           Δ μ = 0.3  10⁻³ kg / m

8)

           μ = (2.3 ±0.3)  10⁻³ kg / m

8 0
3 years ago
Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 4.50 ✕ 10
vovangra [49]
Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.
6 0
3 years ago
A block of plastic in the shape of a rectangular solid that has height 8.00 cm and area A for its top and bottom surfaces is flo
kifflom [539]

Answer:

0.35 kg

Explanation:

8 cm = 0.08 m

For the block to stay balance, the buoyancy force must be the same as gravity that pulls it down.

Let mass of the block be M, then the gravity would be Mg

Let water density be \rho_w = 1000 kg/m^3, the buoyancy force would be the weight of water that is displaced by the submerged block.

For example, when there is no coin, block is h_0 = 0.0312m submerged. The weight of water displaced must be

W_0 = Ah_0\rho_wg = 0.0312A1000g = 31.2Ag

Which is also the weight of block, of Mg

Therefore M = 31.2A.    (1)

As coins are stacked on top of block, h increase, so as weight of water displaced and total weight of block and coins. Now let m be the total weight of coins. The gravity of block and weight must be (M+m)g. And the weight of water displaced is:

W = Ah\rho g = (M + m)g

h = \frac{M}{A\rho} + \frac{m}{A\rho}

Since the linear plot of h vs m has a slope of 0.089 m/kg, we can interpret it as

\frac{1}{A\rho} = 0.089

A = \frac{1}{0.089\rho} = \frac{1}{89} = 0.011 m^2

So from the eq. (1) we can solve for M = 31.2A = 0.35 kg

8 0
3 years ago
At 20 degrees Celsius, conducting wires made of different materials have the same length and the same diameter. Which wire has t
DENIUS [597]

Answer:

Aluminium

Explanation:

Aluminium has the least resistance since It has 3 free electrons per atom. Its resistivity is low compared to other metals provided in the choices (gold, nichrome, tungsten). Low resistivity of metals means a high conductance of the metal referred to.

7 0
4 years ago
Sodium = Na
ad-work [718]
Boron - B
Explanation
6 0
4 years ago
Read 2 more answers
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