Pick the correct word to fill up the blank in below question: velocity, combustion, acceleration, gravitation, force

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an

liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = $\frac{I^2}{R}$ ........1

put here value we get

Power consumed = $\frac{16.7^2}{7.2}$

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0

3 years ago

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A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold

lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

fr - Wₓ = m a (1)

Y axis

N- $W_{y}$ = 0

N = W_{y}

let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

the friction force has the formula

fr = μ N

fr = μ Wy

fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

F = fr

F = (μ mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0

3 years ago

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

PSYCHO15rus [73]

Answer:

Energy density will be 14.73 $J/m^3$

Explanation:

We have given capacitance $C=225\mu F=225\times 10^{-6}F$

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm $0.2\times 10^{-3}m$

We know that there is relation between electric field and potential

$E=\frac{V}{d}$ , here E is electric field, V is potential and d is separation between the plates

So $E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C$

Energy density is given by $E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3$

5 0

2 years ago

If a total force exerted by water in a container with a bottom area of 2 square meters is 450 newtons, what's the water pressure

m_a_m_a [10]

its B 0.225kPa using the formula p=f/A then change the pascals into kpa

7 0

3 years ago

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