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Anuta_ua [19.1K]
3 years ago
10

Mars passes closer to earth than jupiter does, but with an earth-based telescope it is easier to see details on jupiter than on

mars. why is this?
Physics
1 answer:
nalin [4]3 years ago
6 0
Through an earth-based telescope you can see Jupiter more clearly than Mars because Jupiter is much larger than Mars and through the telescope you see the larger objects more clearly.
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calculate the mass of displaced water when a piece of 30cm Iceberg with surface area 1000 cm^2 floats on water density of ice is
sweet [91]

Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.

Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.

27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.

3 0
2 years ago
Which of the following layers in the earth has the highest density?
Ray Of Light [21]
If we are being specific, the inner core has the highest density, but if not then the core in general
5 0
3 years ago
A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
8090 [49]

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0
2 years ago
When two objects collide their momentum after the collision is explained by what?
Gnesinka [82]
<span>When two objects collide their momentum after the collision is explained by</span> the conservation of momentum
5 0
3 years ago
A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti
vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

5 0
3 years ago
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