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3 years ago

Apparent brightness depends on which two properties?

Physics

2 answers:

Serjik [45]3 years ago

6 0

Answer:

I believe it is luminosity and distance

Explanation:

So B

hodyreva [135]3 years ago

4 0

Brightness-Luminosity Relationship:

At a particular Luminosity, the more distant an object is, the fainter its apparent brightness becomes as the square of the distance
So the correct ans is B

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You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kg

Evgen [1.6K]

Answer:

765,000Joules or 765kJ

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

Q = 765,000

Hence the amount of energy needed is 765,000Joules or 765kJ

8 0

3 years ago

Can anyone help me please

ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

Explanation:

Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0

3 years ago

5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3.

user100 [1]

Answer:

$1.04\times 10^7\ J.$

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, $W=P\times\Delta V$

Pressure, P=1.01 × 105 Pa.

Final volume, $V_f=8.50\ m^3.$

Initial volume, $V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.$

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, $Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.$

Also, according to First law of thermodynamics:

$\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.$

Hence, this is the required solution.

8 0

3 years ago

If an object is thrown upward with an initial velocity of 128 ft/second, then its height after t seconds is given by the follow

IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

$h = 128t-32t^2$ ....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

$\dfrac{dh}{dt}=0$

From equation (I)

$\dfrac{dh}{dt}=128-64t$

$128-64t=0$

$t=\dfrac{128}{64}$

$t=2\ sec$

Now, for maximum height

Put the value of t in equation (I)

$h =128\times2-32\times4$

$h=128\ ft$

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

$128t-32t^2=0$

$t(128-32t)=0$

$128=32t$

$t=4\ sec$

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0

3 years ago

Who was the discoverer of the neutron

Pavel [41]

Answer:

James Chadwick

Explanation:

In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron

6 0

2 years ago

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