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2 years ago

What is the quantity of work done when a crane lifts a 100-n block from 2 m above the ground to 6 m above the ground?.

1 answer:

7 0

Answer: by definition work = Force * distanceForce = 100Ndistance = 6 - 2 = 4work = 100 * 4 = 400 J or CStill stuck? Get 1-on-1 help from an expert tutor now.

Explanation:

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An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

kodGreya [7K]

Answer:

about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

= (22.8 ± √823.64)/(-9.8)

The positive solution is ...

t ≈ 0.60195193

It takes about 602 milliseconds for the arrow to reach the ground.

8 0

3 years ago

Which example best illustrates kinetic energy?

mihalych1998 [28]

Answer:

Explanation:

In all actuality. The car may not be fully moving but since it's on a hill and the earth moves it makes the most sense. Also since the car is at a slope by the law of motion what goes up must come down so the car is in fact moving.

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3 years ago

Read 2 more answers

I WILL GIVE BRAINLIEST!!!!

MAVERICK [17]

Answer:

His machine must increase

Explanation:

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3 years ago

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What generalization can be made about energy?

HACTEHA [7]

Well, energy can be produced from the burning of fossil fuels which produce steam to spin turbines which generates electricity. But fossil fuels like coal and oil are no renewable.
Some renewable ways to get energy is water where they separate hydrogen from h2o because hydrogen can be made into energy. Also nuclear fission is another way to get energy. That's when in a nuclear power plant, the atoms nucleus is split which sends a large amount of energy.
Wind energy is another source of energy, which is from windmills. Another renewable energy source is solar energy. That's when u use solar cells, like some houses and calculators have, which generates a lot of energy, but only in the day since the sun doesn't shine in the nigh.

Hope I helped!!!

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3 years ago

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a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago