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3 years ago

How are climate and weather related?

1 answer:

8 0

Weather is what conditions of the atmosphere are over a short period of time, and climate is how the atmosphere "behaves" over relatively long periods of time.

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9. What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building? (Each storey is

kow [346]

Answer:

15009

Explanation:

PE = mgh

PE = 61.2(9.81)(10 * 2.50)

PE = 15009.3

3 0

2 years ago

An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?

zubka84 [21]

The force needed to accelerate an elevator upward at a rate of $2 m / s^{2}$ is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as $2 m / s^{2}$ , the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

$\text {Force}=\text {Mass} \times \text {Acceleration}$

$\text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }$

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of $2 m / s^{2}$ .

3 0

3 years ago

Doubling the frequency of a wave source doubles the speed of the wave assuing that wavelength remains the same? True or false? W

SVETLANKA909090 [29]

Answer:true,because velocity is directly proportional to speed or velocity

Explanation:

Velocity = frequency x wavelength

The velocity or speed varies directly with the frequency, so as the frequency is increased, the velocity or speed is also increased

8 0

3 years ago

If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th

olga55 [171]

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

3 0

3 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago