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CaHeK987 [17]
3 years ago
11

Can anyone pls help me with this I’ll appreciate it. I’m trying to study for a test but I don’t know the answer to this.

Physics
1 answer:
forsale [732]3 years ago
7 0

Answer:

okay so what you will do is where is says red giant you will write all about what it talks about red giants only, and the box plantary nebulas you will write about what is says about only planetary nebulas. x- hope this helps :)

Explanation:

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A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?
kompoz [17]

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

7 0
3 years ago
A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the
MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

v^{2}=u^{2}+2as and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

u=\sqrt {v^{2}-2gs}

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s

5 0
4 years ago
Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At w
Alisiya [41]

Answer:

C. At the instant the ball reaches its highest point.

Explanation:

When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"

Remember, F = ma = m(v/t)

Since v = 0 at maximum height

F = m(0/t)

F = 0N

This shows that the force acting on the body is zero at the maximum height.

4 0
3 years ago
?during the first few days of a fast, what energy source provides about 90% of the glucose needed to fuel the body?
Kobotan [32]
<span>The energy source provides about 90% of the glucose needed to fuel the body in the first few days of the fast is protein. protein is a string of amino acids that peforms a variety of functions to the body. These are found in foods like eggs, cheese, chicken breasts, meat and many more/.</span>
8 0
3 years ago
A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l
son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Using formula of focal length for glass ethyl alcohol system

\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Divided equation (II) by (I)

\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}

Where, n_{g} = refractive index of glass

n_{a} = refractive index of air

n_{ethyl} = refractive index of ethyl

Put the value into the formula

\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}

\dfrac{f'}{11.5}=\dfrac{25}{7}

f'=\dfrac{25}{7}\times11.5

f'=41.07\ cm

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0
3 years ago
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