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3 years ago

How much energy to be removed to drop the temperature of 5.7kg of wood from 20degrees to 7degrees. #100points

2 answers:

5 0

see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk

WITCHER [35]3 years ago

5 0

Answer:

D is correct they were correct

Explanation:

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How do paved parking lots and roads with concrete or asphalt affect water flow on the land?

harina [27]

On the regular ground, water usually seeps in to the soil, which has a number of benefits for the wildlife in that area. However, with asphalt and concrete, there is no soft and absorbent soil to take in the water, so it just keeps flowing down to the lowest part of land it can. Additionally, some water runoff can carry fertilizer and other harmful chemicals with it in to the oceans and lakes it’s dumped in, which harms the ecosystem in them as well.

8 0

3 years ago

How does the kinetic energy from the forward motion of a car traveling at 16 m/s

elena55 [62]

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

For the first case, velocity v = 16m/s
For the second case, velocity = 8m/s
Let the mass of the car be m

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

7 0

2 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

blondinia [14]

Answer:

$\theta = 20.98 degree$

Explanation:

As we know that the speed of the sound is given as

$v = 332 + 0.6 t$

now at t = 273 k = 0 degree

$v = 332 m/s$

so we have

$a sin\theta = N\lambda$

$a sin\theta = N(\frac{v_1}{f})$

now when temperature is changed to 313 K we have

$t = 313 - 273 = 40 degree$

now we have

$v = 332 + (0.6)(40)$

$v_2 = 356 m/s$

$a sin\theta' = N(\frac{v_2}{f})$

now from two equations we have

$\frac{sin19.5}{sin\theta} = \frac{332}{356}$

so we have

$sin\theta = 0.358$

$\theta = 20.98 degree$

7 0

3 years ago

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

5 0

3 years ago