Which of the following best describes the electromagnetic spectrum?

An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around whic

Cloud [144]

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

$a_{t}=r\alpha$

$a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}$

$a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}$

$a_{t}=0.0301\ m/s^2$

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

4 0

3 years ago

You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of th

solmaris [256]

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

$v_x = u cos \theta$

where

u = 8.3 m/s is the initial velocity of the balloon

$\theta=27^{\circ}$ is the angle of projection

Substituting,

$v_x = (8.3)(cos 27^{\circ})=7.4 m/s$

b)

The vertical component of the initial velocity of a projectile is given by

$u_y = u sin \theta$

where

u is the initial velocity

$\theta$ is the angle of projection

Here we have

u = 8.3 m/s

$\theta=27^{\circ}$

Substituting,

$u_y = (8.3)(sin 27^{\circ})=3.8 m/s$

c)

The vertical component of the velocity of the balloon follows the suvat equation

$v_y = u_y - gt$

where

$v_y$ is the vertical velocity at time t

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

$v_y = 0$

So we get:

$0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s$

d)

The maximum height of the balloon can be calculated using the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

$s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m$

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

$t=\frac{2u_y}{g}$

where

$u_y$ is the initial vertical velocity

$g$ is the acceleration of gravity

Here we have

$u_y = 3.8 m/s$

$g=9.8 m/s^2$

Substituting,

$t=\frac{2(3.8)}{9.8}=0.78 s$

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

Here we have

$v_x = 7.4 m/s$

$t = 0.78 s$

Substituting,

$d=(7.4)(0.78)=5.77 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

Ask Your Teacher An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires

natta225 [31]

Answer:

The potential difference at the customer's house is 117.1 V.

Explanation:

a) The potential difference at the customer's house can be calculated as follows:

$\Delta V_{h} = \Delta V_{p} - \Delta V_{l}$

<u>Where</u>:

$V_{h}$ : is the potential difference at the customer's house

$V_{p}$ : is the potential difference from the main power lines = 120 V

$V_{l}$ : is the potential difference from the lines

$\Delta V_{h} = \Delta V_{p} - IR$

The resistance, R, is:

$\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega$

Now, the potential difference at the customer's house is:

$\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V$

Therefore, the potential difference at the customer's house is 117.1 V.

I hope it helps you!

4 0

3 years ago

Why are cockpits not pressurized

Darina [25.2K]

It be more productive to just have the pilot were an oxygen mask instead of using the entire section for air when the pilot is siting down. The glass may also fog up, causing the entire glass pane to explode at high altitude.

5 0

3 years ago

Sunlight is a form of electromagnetic energy.

Sergio039 [100]

Answer: True

Explanation: Hope this helps :)

7 0

3 years ago