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3 years ago

Which object has the HIGHEST mass?

1 answer:

5 0

I think it would be A: gold ring

if you were to place all of these on a bed of water, the ring would be the first to sink.

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Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

• 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is

Kipish [7]

Answer:

0.5 m/sec

Explanation:

v=S÷t

=5÷10

v=0.5 m/sec

6 0

3 years ago

A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

3 years ago

An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet

zimovet [89]

Answer:

F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v²

Em₀ = Emf

e ΔV = ½ m v²

v =√ 2 e ΔV / m

v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

v = √(1.8075 10¹⁶)

v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Emo = U = q DV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v2

Emo = Emf

.e DV = ½ m v2

.v = RA 2 e DV / m

.v = RA (2 1.6 10-19 51400 / 9.1 10-31)

.v = RA (1.8075 10 16)

.v = 1,344 108 m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10-19 1,344 108 0.4

F = 8.6 10-12 N

5 0

3 years ago

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

valkas [14]

Op here is another problem exactly like that. Just plug in your variables instead. And remember, time is never negative.

5 0

3 years ago