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Brilliant_brown [7]
2 years ago
13

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters

. What is the change in the potential energy (in Joules) of the mass as it goes up the incline?

Physics
1 answer:
Tanya [424]2 years ago
5 0

Answer:

The change in potential energy of the mass as it goes up the incline is 0.343 joules.

Explanation:

We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:

U_{g}= m\cdot g\cdot \Delta y

Where:

U_{g} - Gravitational potential energy, measured in Joules.

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

\Delta y - Change in vertical height, measured in meters.

This work is the energy needed to counteract effects of gravity at given vertical displacement.

If we know that m = 0.5\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta y = 0.07\,m, the change in the potential energy of the mass as it goes up the incline is:

U_{g} = (0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.07\,m)

U_{g} = 0.343\,J

The change in potential energy of the mass as it goes up the incline is 0.343 joules.

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nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
2 years ago
7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the
exis [7]

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

distance (r) = 210 m

intensity (I) = power/area

where we assume the area of a sphere due to the uniformity of the output in all directions

area = 4πr^{2} =  4π x 210^{2} = 554,176.9 m^{2}

intensity (I) = \frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}

therefore the desired ratio I/I0 = \frac{113.7 x 10^{-12}}{1 x 10^{-12}} = 113.7

7 0
2 years ago
In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p,
viktelen [127]

Answer:

Explanation:

In case of gas , work done

W = ∫ p dV , p is pressure and dV is small change in volume

the limit of integration is from Vi to Vf .

= ∫ p dV

=  ∫ p₀V^{-\frac{6}{5}  dV

= p₀ V^{-\frac{6}{5} +1} / ( \frac{-6}{5} +1 )

=  - 5p₀ V^{-\frac{1}{5}

Taking limit from Vi  to Vf

W = - 5 p₀ ( V_f^\frac{-1}{5} - V_i^{\frac{-1}{5}  ) ltr- atm.

7 0
2 years ago
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
2 years ago
Read 2 more answers
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