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Alekssandra [29.7K]

4 years ago

12

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

.97 x 1024 kg, radius of the earth = 6.37 x 106 m)

1 answer:

Neporo4naja [7]4 years ago

8 0

Answer:

b

Explanation:

6kseld;r78t0gvyu79bmkp,l[;.l,kjmnhbgyvft bujhnkmlokjnhiubgyfv vaxjosnihdfbamlskjfdnh

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m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resisto

notsponge [240]

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

<u>T₂ = 49.3°C</u>

8 0

3 years ago

A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at

Liono4ka [1.6K]

Answer:

(a) the runner's kinetic energy at the given instant is 308 J

(b) the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

$K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J$

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

$K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J$

the change in the kinetic energy is calculated as;

$\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4$

Thus, the kinetic energy increased by a factor of 4.

4 0

3 years ago

Why is it not advisable to sterilize clinical thermometer with boiling water

amm1812

It will break. A clinical the emoter is designed to measure relatively small changes in temperature. The substance inside a clinical thermoset will expand or evaporate at boiling temperatures and break.

6 0

3 years ago

A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5

Alina [70]

Here we go.
My abbreviations; KE = Kinetic Energy; GPE = Gravitational Potential Energy.

So first off, we know the fish has KE right when the bird releases it. Why? Because it has horizontal velocity after released! So let’s calculate it:
KE = 1/2(m)(V)^2
KE = 1/2(2)(18)^2
KE = 324 J

Nice!
We also know that the fish has GPE at its maximum height before release:
GPE = mgh
GPE = (2)(9.81)(5.40)
GPE = 105.95 J

Now, based on the *queue dramatic voice* LAW OF CONSERVATION OF ENERGY, we know all of the initial energy of the fish will be equal to the amount of final energy. And since the only form of energy when it hits the water is KE, we can write:
KEi + GPEi = KEf
(Remember - we found the initial energies before!)
(324) + (105.95) = KEf
KEf = 429.95J
And that’s you’re final answer! Notice how this value is MORE than the initial KE from before (324 J) - this is because all of the initial GPE from before was transformed into more KE as the fish fell (h decreased) and sped up (V increased).

If this helped please like it and comment!

4 0

3 years ago

An object at rest is suddenly broken apart into fragments a and b by an explosion. the fragment a acquires three times the kinet

lbvjy [14]

Momentum will be conserved in one dimension in the explosion.

<span> Given that the fragment a acquires three times the kinetic energy of the fragment b.
<span>
P</span><span><span>initial </span><span>= p</span></span>final ⇒ 0 =mₐv⁰ₐ+mьv⁰ь= 0 ⇒ v⁰ь = -mₐv⁰ₐ/mь

KE= 3KEь

⇒1/2 mₐv⁰ₐ² = 3 (1/2mьv⁰ь²)
</span><span>
⇒1/2 mₐv⁰ₐ² = 3/2 mь(-mₐv⁰ₐ/mь)²

⇒1/2 mₐv⁰ₐ² = 3/2 mь(mₐ²v⁰ₐ²/mь²)

</span>

⇒1/2 x 2/3 = mₐ/mь= 1/3

<span> <span> Thus the ratio of the masses of the fragments is 1:3. </span></span>

4 0

4 years ago