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dmitriy555 [2]
2 years ago
13

HELP PLS! (LOOK AT THE PICTURE)

Physics
1 answer:
Arturiano [62]2 years ago
4 0
The answer is b !!!! Hope it helps
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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
3 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
3 years ago
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i
ELEN [110]

Answer:

Pls see attached file

Explanation:

5 0
3 years ago
La mascota de felipe esta jugando en el parque despues de un tiempo esta agotado de tanto correr. Felipe conoce que el perro con
MatroZZZ [7]

Responder:

6.704 m / s

Explicación:

Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.

Trabajo realizado = Fuerza × distancia

Distancia = Trabajo realizado / Fuerza

Distancia = W / mg

S = 176/8 × 9,81

S = 176 / 78,48

S = 2,24 m

Dada la velocidad inicial u = 3.6km / h

Convertir a m / s

= 3.6km × 1000m / 1h × 3600

= 3600/3600

= 1 m / s

u = 1 m / s

Usando la ecuación de movimiento

v² = u² + 2gS para obtener la velocidad final v:

v² = 1² + 2 (9,81) (2,24)

v² = 1 + 43,9488

v² = 44,9488

v = √44,9488

v = 6,704 m / s

Por tanto, la rapidez final del perro es de 6,704 m / s

6 0
2 years ago
Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i
astra-53 [7]
The average speed would be 33.29m/s.
The average speed equation is:

Average speed =  \frac{total distance}{total time}

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled. 

Scenario 1:
Speed = 29m/s
Time = 120s
Distance = ?

Distance = (29m/s)(120s)
               = 3,480m

Scenario 2
Speed = 35m/s
Time = 300s
Distance = ? 

Distance = (35m/s)(300s)
               = 10,500m

Now that you have the distance of both, you can solve for your average speed. 

Average speed = \frac{total distance}{total time}
                                = \frac{3,480m+10,500m}{120s+300s}
                                = \frac{13,980m}{420s}
                                = 33.29m/s
5 0
3 years ago
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