You had an amazing race. You finished the 300kms race in 1hr 40 minutes. What is the average velocity?

HELP ME ASAP ITS PHYSICS AND it’s finding the kinetic energy SHOW WORKKK

lord [1]

k = 1/2 × m × v^2

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1. k = 1/2 × 60 × ( 10 )^2

k = 30 × 100

k = 3000 j

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2. k = 1/2 × 60 × ( 9 )^2

k = 30 × 81

k = 2430 j

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3. k = 1/2 × 60 × ( 4 )^2

k = 30 × 16

k = 480 j

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4. k = 1/2 × 60 × ( 6 )^2

k = 30 × 36

k = 1080 j

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5. k = 1/2 × 60 × ( 0 ) ^2

k = 30 × 0

k = 0 j

6 0

3 years ago

Read 2 more answers

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through

Irina18 [472]

Answer:

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .

Let in this last leg , u be the initial velocity.

s = ut + 1/2 g t²

2 = .2 u + .5 x 9.8 x .04

u = 9.02 m /s .

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02)² + 2 x 9.8 x 2

= 81.36 +39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped . Let it be h.

Initial velocity u = 0

v² = u² +2gh

(10.97 )² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m

5 0

4 years ago

You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you

givi [52]

Answer:

4) True, the distance was reduced by a de factor 2

Explanation:

To analyze the statements, let's first see how the sound is produced

The cricket makes the sound rubbed its legs, the energy needed to produce it is constant, after the sound is produced it expands in all directions and this wave is subject to the law of energy conservation.

By expanding in all directions we can assume that the energy for the time that is the power of the sound is distributed over a sphere with different radii each time. The perceived intensity is defined as the power per unit area

I = P / A

Where P is the power, A are the areas and I are the intensities heard by the person

Let's clear the power and match, for two different radii (distances)

P = I₁ A₁ = I₂ A₂

The area of a sphere is

A = 4π R₂

I₂ = I₁ A₁ / A₂

I₂ = I₁ 4π R₁² / 4π R₂²

I₂ = I₁ R₁² / R₂²

Let's analyze this last equation for the situation presented if the intensity increases 4 times, substitute

I₂ = 4 I₁

4 I₁ = I₁ R₁² / R₂²

R₂ = R₁ / 2

We see that to increase the intensity 4 times the distance was reduced twice

Now we can analyze the expressions given

2) False, the power is constant, the cricket always does the same job

3) False, the speed of the wave is constant in the air

4) True, the distance was reduced by a de factor 2

3 0

3 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

Match the term and the definition

sineoko [7]

Answer:

1. When a cold air mass and a warm air mass meet but are at a standstill, the boundary is called stationary front.

2. If warm or more humid air presses forward and colder or drier air draws back, the boundary is called warm front.

3. If colder or drier air presses forward and warmer or more humid air draws back, the boundary is called cold front.

4. Sometimes a warm air mass can get caught between two cold fronts, which can force the warm air up into a wedge shape is called occluded front.

3 0

3 years ago