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Misha Larkins [42]

1 year ago

14

A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i

s 45 N; therefore, the magnitude of the force of tension is
N.

1 answer:

lukranit [14]1 year ago

8 0

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

nnnn

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

1 year ago

. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

faust18 [17]

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

$h = gt^2/2$

$10 = 10t^2/2$

$t^2 = 2$

$t = \sqrt{2} = 1.414 s$

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

$\theta = \alpha t^2/2$

$\theta = 15 * 2/2 = 15 rad$

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0

2 years ago

Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There

mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π) ΔV² / r⁴

Explanation:

a) For this exercise we can use Coulomb's law

F = - k Q² / r²

where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

Capacitance is defined by

C = Q / ΔV

Q = C ΔV

also the capacitance for a parallel plate capacitor is related to its shape

C = ε₀ A / r

we substitute

Q = ε₀ A ΔV / r

we substitute in the force equation

F = k (ε₀ A ΔV / r)² / r²

k = 1 / 4πε₀

F = ε₀ /4π L² ΔV² / r⁴4

F = L² ΔV² ε₀/ (4π r⁴)

F = (L² ε₀/4π) ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

F = q E

where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface

Ф = ∫E .dA = $q_{int}$ / ε₀

the plate have two side

2E A = q_{int} / ε₀

E = σ / 2ε₀

σ = q_{int} / A

substituting in force

F = q σ / 2ε₀

the charge total on the other plate is

q = σ A

q = σ L²

F= σ² L² /2ε₀

4 0

2 years ago

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.

5 0

1 year ago

Read 2 more answers

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

1 year ago