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4 years ago

Why is it important to use constants in an experiment?

1 answer:

5 0

<span>In order for the results to be valid, the dependent variable can only be affected by the independent variable, so somethings need to be kept constant. The things that need to be kept constant are called controlled variables.</span>

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Which of the following phenomena provides evidence of electric fields in the atmosphere?

frosja888 [35]

The ground(Earth's surface) is at low potential whereas the clouds are at higher potential. Due to this potential difference, we see lightning.

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3 years ago

Read 2 more answers

Calcula el peso de una mesaven la tierra cuya masa es de 60 kg

blsea [12.9K]

Answer:

W = 588 N

Explanation:

The given question is ,"calculate the weight of a table on the earth whose mass is 60 kg"

The mass of the table, m = 60 kg

We need to find the weight of a table on the earth.

The weight of an object is given by :

W = mg

Where

g is acceleration due to gravity

W = 60 kg × 9.8 m/s²

W = 588 N

Hence, the weight of the table is 588 N.

5 0

3 years ago

Which statement is a scientific theory?

Dmitry [639]

Answer:

I think it is D

Explanation:

7 0

3 years ago

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

What is the product of mass and velocity called?

Crank

The product of mass and velocity is momentum. P=mv. The 'P' in the equation stands for momentum.

3 0

3 years ago