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3 years ago

Using the work energy theorem. What is the velocity of a 850kg car after starting at rest when 13,000J of work is done to it.

Physics

1 answer:

Misha Larkins [42]3 years ago

3 0

When the car starts moving it acquires kinetic energy.

For this problem the formula of kinetic energy is:

$\frac{1}{2}mv^2$

Where m is the mass of the car = 850kg, and v is the speed of the car.

If we consider insignificant the energy lost by friction with soil and air we can propose the following equation:

Applied energy = Kinetic energy acquired by the car.

$13000\ J = \frac{1}{2}mv^2\\\\13000\ J = \frac{1}{2}(850)v^2\\\\\frac{13000(2)}{850} = v^2\\\\v = \sqrt{\frac{13000(2)}{850}}\\\\v = 5.531 m/s$

The velocity is 5.531 m/s

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Which of the following accurately describes the way in which a muscle moves?

Vaselesa [24]

<h3>Answer;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3>Explanation;</h3>

The events of muscle fiber shortening occurs with in the sacromeres in the fibers.
Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.
Filament movement starts at the region or zone where thin and thick filaments overlap.
Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.

7 0

3 years ago

Por favor, necesito ayuda urgente 4. vectores perpendiculares de 25 y 40 unidades cada uno. Hallar gráfica y numericamente el ve

Darina [25.2K]

Answer:

4) R = 47.17 units , 5) R= 10,29 unidades, 6) R= 2994,4 km , θ = -33,7

Explanation:

Este es un ejercicio de adición de vectores

4) como los vectores son perpendiculares.

Para encontrar la resultante podemos usar el teorema de pitoras

R = √ a² + b²

R = √√ ( 25² + 40²)

R = 47,17 unidades

5) Este caso como el angulo es diferente de 90 debemos usar la relación de pitoras completa

R² = a² + b² + 2 a b cos θ

donde el angulo es entre los vectores a y b

R² = 12² + 16² – 2 12 16 cos 40

R²= 400 – 294,16

R= 10,29 unidades

6) En este caso los dos desplazamientos son perpendiculares, por lo cual Usamos el teorema de Pitágoras

R = √ (2400² + 1600²)

R= 2994,4 km

para el angulo de este desplazamiento usamos trigonometría

tan θ = y/x

θ = tan⁻¹ y/x

θ = tan⁻¹ 1600/(-2400)

θ = -33,7

TRASLATE

This is a vector addition exercise

4) as the vectors are perpendicular.

To find the result we can use the Poreor theorem

R = √ a² + b²

R = √ (25² + 40²)

R = 47.17 units

5) This case, as the angle is different from 90, we must use the complete ratio of the pitoras

R² = a² + b² + 2 a b cos θ

where the angle is between vectors a and b

R² = 12² + 16² - 2 12 16 cos 40

R² = 400 - 294.16

R = 10.29 units

6) In this case the two displacements are perpendicular, which is why we use the Pythagorean theorem

R = √ (2400² + 1600²)

R = 2,994.4 km

for the angle of this displacement we use trigonometry

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 1600 / (- 2400)

θ = -33.7

3 0

3 years ago

Relation of Kilometre and mile

Alja [10]

Answer:

1km = o.621371 mile

Explanation:

1.609 kilometers equal 1 mile. The kilometer is a unit of measurement, as is the mille. However, a mile is longer than a kilometer.

5 0

3 years ago

Read 2 more answers

When water (H₂O) freezes into ice, some of the properties have changed. What stays the same?

erastova [34]

Answer:

1st one

identity of the H₂O

When water (H2O) freezes into ice, some of the properties have changed but the identity of the H2O is the same. Explanation; Water can exist in liquid, solid; ice and gaseous state; steam. Freezing occurs when water is continuously cool until it turns to solid ice, water freezes at zero degree Celsius

Hope This Helps

4 0

2 years ago

A 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same cons

LenKa [72]

Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
$F=ma$
from which we find an expression for the acceleration:
$a= \frac{F}{m}$ (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
$v_f^2 - v_i^2 = 2 a S$ (2)
where
$v_f$ is the final speed of the object
$v_i$ is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing $v_f$ (since the final velocity of the two objects is zero), we find
$-v_i^2 = 2 \frac{F}{m}S$
$S=- \frac{v_i^2 m}{2F}$
where we can ignore the negative sign (because the force F will bring another negative sign).

For the first object, we have
$S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]$
And for the second object we have
$S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]$

And since the braking force applied to the two objects is the same, the two objects cover the same distance.

3 0

4 years ago