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lukranit [14]
2 years ago
15

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

has a speed of 12 m/s. Which of the following correctly identifies whether the object-Earth system is open or closed and describes the net external force?
A. The system is closed, and the net external force is zero.
B. The system is open, and the net external force is zero.
C. The system is closed, and the net external force is nonzero.
D. The system is open, and the net external force is nonzero.
Physics
1 answer:
makkiz [27]2 years ago
4 0

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

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xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
1 year ago
What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i
Katarina [22]

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0
1 year ago
A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.
kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

F_f = \mu N = \mu mg

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s

3 0
1 year ago
Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
Alinara [238K]

The volume of the room is the product of its dimensions:

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Now, from the equation

d=\dfrac{m}{V}

where d is the density, m is the mass and V is the volume, we deduce

m=dV

So, multiply the density and the volume to get the mass of air in the room.

7 0
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tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

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x = final position

x_{o} = initial position of the object at the start of the motion

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