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MrRissso [65]
3 years ago
15

A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas

t!!!)
Physics
1 answer:
Arada [10]3 years ago
7 0

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

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B. +Q, + W are the correct sign

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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

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A piano mover raises a 1000-n piano at a constant speed using a very light rope in a frictionless pulley system, as shown in the
Rina8888 [55]

Here As we can see the figure that the end of the rope is pulled by some force F

Now as we can see that Piano is connected by a pulley which is passing over the pulley so effectively net force on the piano upwards will be 2F as it is connected by 2 ropes by the pulley

Now for constant velocity of the piano we will say

2F - mg = ma

since velocity is constant so acceleration must be ZERO

so here we have

2F - mg = 0

as we know here that

mg = 1000 N

so we will have

2F - 1000 = 0

F = 500 N

so here force must be 500 N

5 0
4 years ago
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B. The silica cylinder of a radiant wall heater is 0.6 m long
SIZIF [17.4K]

So,  If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

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  • A = surface area of cylindrical wall heater = 2πrh where
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Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π  × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72  × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

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