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3 years ago

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Explain your results: How can Earth’s gravity affect the water when the water isn’t actually touching the Earth? (2 pts) Do you

think the Moon’s gravity is affecting the water in the beaker? Why or why not? (2 pts) If you were on the Moon, do you think you would feel the pull of Earth’s gravity? Explain! (2 pts)

1 answer:

Levart [38]3 years ago

5 0

Answer:

Because the Earth has so much gravity, it can hold water, land, and life in it's atmosphere.

(Not sure what beaker you are talking about, so sorry) But I don't think the moon's gravity would have an effect on a beaker of water because the Earth's gravity is much more than the moon's.

I think you would be able to feel a little bit of Earth's gravity on the moon because the Earth's gravity pulled the moon into orbit, therefore, gravity on Earth my have some effect on the moon.

hope this helps!

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An unknown substance has a mass of 11.9 g . When the substance absorbs 1.071×102 J of heat, the temperature of the substance is

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Answer: The most likely identity of the substance is iron.

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $1.071\times 10^2$ Joules

m= mass of substance = 11.9 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$ = 25.0°C

Final temperature of the water = $T_f$ = 45.0°CChange in temperature , $\Delta T=T_f-T_i=(45-25)^0C=20^0C$

Putting in the values, we get:

$1.071\times 10^2=11.9\times c\times 20^0C$

$c=0.45J/g^0C$

The specific heat of 0.45 is for iron and thus the substance is iron.

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It would appear as blue

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P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te

tankabanditka [31]

Answer:

(a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

$kx-mg=ma$

$x=\dfrac{ma+mg}{k}$ ....(I)

The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

$x=2.5\times x_{0}$

Here, acceleration is zero

So, $x=2.5\times\dfrac{mg}{k}$

We need to calculate the acceleration

Put the value of x in equation (I)

$2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}$

$2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)$

$a=2.5g-g$

$a=1.5g$

Hence, (a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

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What is the relationship between thickness of lens and focal length?

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Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

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An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

Ipatiy [6.2K]

Given that,

Time = 0.5 s

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Using equation of motion

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Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

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(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times10\times(0.5)^2$

$s=1.25\ m$

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

$v_{avg}=\dfrac{\Delta x}{\Delta t}$

$v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}$

Where, $x_{f}$ = final position

$x_{i}$ = initial position

Put the value into the formula

$v_{avg}=\dfrac{1.25+0}{0.5}$

$v_{avg}=2.5\ m/s$

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

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