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max2010maxim [7]

1 year ago

15

According to newton, doubling the distance between two interacting objects:

1 answer:

Furkat [3]1 year ago

8 0

Answer:

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).

Explanation:

hope his helps

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In the Bohr model of the atom, an electron in an orbit has a fixed ____.

worty [1.4K]

The answer is C. an electron in an orbit has a fixed energy.

3 0

2 years ago

A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must

umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0

2 years ago

A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km

Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

$v_f=v_o+at$

Since the car starts at rest, you can write the following equation:

$20=0+a(5) \\\\a=20\div 5=4 m/s^2$

Now that you have the acceleration, you can do this:

$d=v_o+\dfrac{1}{2}at^2$

Once again, there is no initial velocity:

$d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m$

Hope this helps!

8 0

3 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

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2 years ago

BABY BABY BABY OOOHHHH I THOUGHT THAT U WOULD ALWAYS BE MINE

kenny6666 [7]

Answer:

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Explanation:

6 0

2 years ago

Read 2 more answers