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malfutka [58]
3 years ago
10

A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 deg

rees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.
Physics
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

Explanation:

energy stored in spring initially

= kinetic + potential energy of block + energy dissipated by friction

= 1/2 mv² + mgh + μ mgcosθ x  d

m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane

= m (1/2 v² + gh + μ gcosθ x  d )

= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )

= 1.05 ( 13.005 + 27.543 + 21.635)

= 65.3 J .

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A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

                      = (14) (9.81) (5.5sin36.9°)

                       = 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

                      = -453.5J

8 0
3 years ago
What is the main reason for using a data table for collecting data?
Luba_88 [7]

Answer:

to make calculation more easy to get

Explanation:

if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic

7 0
3 years ago
What is the volume of the cone?
dezoksy [38]

Answer:

42.417 cm³

Explanation:

The formula to find the volume of a cone is :

V = \frac{1}{3} × π r² h

Here,

r ⇒ radius ⇒ 3 cm

h ⇒ height ⇒ 4.5 cm

<u>Let us find it now.</u>

V = \frac{1}{3} × π r² h

V = \frac{1}{3} × π × 3 × 3 × 4.5

V = \frac{1}{3} × π × 9 × 4.5

V = \frac{1}{3} × π × 9 × 4.5

V = \frac{1}{3} × π × 40.5

V = \frac{1}{3} × 3.142 × 40.5

V = \frac{1}{3} × 127.251

V = <u>42.417 cm³</u>

4 0
2 years ago
Which type of evidence was used to tie Ted Bundy to the murder of Lisa Levy?
Elodia [21]

Explanation:

identified by one of the Chi Omega survivors as the assailant, and hair and fiber evidence tied him to the Leach murder. His trial for the Florida attacks in June 1979 was the first-ever nationally televised trial in the United States.

6 0
1 year ago
A coil with an inductance of 2.8 H and a resistance of 12 Ω is suddenly connected to an ideal battery with ε = 89 V. At 0.086 s
Thepotemich [5.8K]

Answer:

The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.

Explanation:

Given that,

Inductance = 2.8 H

Resistance = 12 Ω

Potential \epsilon_{0}=89\ V

Time = 0.086 s

(a). We need to calculate the energy stored in the magnetic field

Using formula of current

i=i_{max}(1-e^(\frac{-t}{\tau}))

Using formula of energy

U=\dfrac{1}{2}Li^2

On differentiating

\dfrac{dU}{dt}=Li\frac{di}{dt}

\dfrac{dU}{dt}=L\dfrac{d}{dt}(i_{max}(1-e^(\frac{-t}{\tau}))

Again differentiating

\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})e^{\frac{-t}{\tau}}

\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-\t\times R}{L}})e^{\frac{-t\times R}{L}}

Put the value into the formula

\dfrac{dU}{dt}=\dfrac{(89)^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})e^{\dfrac{-0.086\times12}{2.8}}

\dfrac{dU}{dt}=140.7\ watt

(b). We need to calculate the thermal energy

Using formula of thermal energy

P=i^2R

P=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})^2

Put the value into the formula

P=\dfrac{89^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})^2

P=62.7\ Watt

(c). We need to calculate the delivered energy by the battery

Using formula of energy

P'=P+\dfrac{dU}{dt}

P'=62.7+140.7

P'=203.4\ watt

Hence, The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.

5 0
3 years ago
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