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2 years ago

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According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme

nt?

1 answer:

Bas_tet [7]2 years ago

8 0

The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.

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Can someone please answer this, ill give you brainliest Would be very appreciated.

marshall27 [118]

Answer:

cohesive properties

Explanation:

The property of cohesion allows liquid water to have <u>no tension on the surface</u>.

7 0

1 year ago

Read 2 more answers

Does heat flow towards cold?

Liula [17]

Answer:

And unless people interfere, thermal energy — or heat — naturally flows in one direction only: from hot toward cold. Heat moves naturally by any of three means. The processes are known as conduction, convection and radiation. Sometimes more than one may occur at the same time.

4 0

2 years ago

Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the

Marrrta [24]

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised from T to T'.

Hence specific heat of a substance is calculated as-

$c= \frac{Q}{m[T'-T]}$

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

7 0

2 years ago

Read 2 more answers

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

just olya [345]

Force is defined as the rate of change of momentum.
The initial amount of momentum is $mv$ because water stops when it hit the wall total change of momentum must be $\Delta p=mv$ .
Now let's calculate the force.
$F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v$
We need to find $\frac{dm}{dt}$ . This is the amount of water hiting the wall per second.
$\frac{dm}{dt}=\rho Av$
Our final formula would be:
$F=\rho Avv=\rho Av^2$
And now we can calculate the answer:
$F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N$

6 0

2 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