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Vsevolod [243]
2 years ago
15

A rock is dropped from a helght of 2.5 meters. What is its velocity when it reaches the ground?

Physics
1 answer:
Nataly_w [17]2 years ago
7 0

u=0 m/s

h=2.5m

Using kinematic eq:

{v}^{2}  -  {u}^{2}  = 2gh \\  {v}^{2}  -  {0}^{2}  = 2gh \\ v =  \sqrt{2gh}  \\ v =  \sqrt{2 \times 10 \times 2.5}  \\ v =  \sqrt{50}  \\ v = 7.07m/s

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A light spring has a force constant of 70 N/m and is used to pull a 10 kg box on a horizontal frictionless surface. If the box h
Sergio039 [100]

Answer:

82.4 cm

Explanation:

∑F = ma

kx cos θ = ma

x = ma / (k cos θ)

x = (10 kg) (5 m/s²) / (70 N/m cos 30.0°)

x = 0.824 m

x = 82.4 cm

8 0
3 years ago
There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid
OLga [1]
To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.
5 0
3 years ago
A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the d
timofeeve [1]
Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²

Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m

Horizontal distance:
s = ut
= 3 x 4
= 12 m

Total distance = √(12² + 16²)
= 20 m.
3 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
Subject- Science <br>Grade- 6th<br>please help me with hw. Thank you!<br>​
velikii [3]

Answer:

b. they get blown in from colder or warmer areas.

3 0
1 year ago
Read 2 more answers
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