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2 years ago

How much does it cost to leave a porch light on all night?

1 answer:

6 0

Cost = (0.001) x (the wattage of the light) x (the number of hours it's left on) x (the cost of each kilowatt-hour of electrical energy where you live).

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A car is moving at 14 m/s. After 30 s, its speed inncreased to 20 m/s. What is the acceleration over time

Juli2301 [7.4K]

Answer:

0.2 m/s^2

Explanation:

initial speed 14m/s

final speed 20m/s

acceleration:

(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2

7 0

2 years ago

What is 62.125 kg rounded to two significant figures?

Mamont248 [21]

 62.125

= 62.12

THANK you 

6 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Ilia_Sergeevich [38]

Below is the attachment of the solutions.

5 0

3 years ago

Help please

Naddik [55]

Answer:

acid and bases

Explanation:

4 0

2 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

2 years ago