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2 years ago

An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?

Physics

1 answer:

Mila [183]2 years ago

7 0

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

$\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s$

Now we can calculate the velocity of a point on the tip of the hand by using the formula

$v=\omega r$

where

$\omega=0.105 rad/s$ is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

$v=(0.105)(0.15)=0.0158 m/s$

Learn more about angular motion here:

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The instantaneous emf resulting from magnetic induction equals the rate of change of flux is

katen-ka-za [31]

Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.

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2 years ago

Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. this can be represented by the fun

Romashka [77]

The values of t are <u>4.643 second</u> for the function $H(t)=-16t^2+73t+6$

What is batter misses?

An out in baseball happens when the umpire declares a batter or baserunner out. A hitter or runner who is out is no longer able to score runs and must go back to the dugout until their subsequent turn at bat. The batting team's turn is over after three outs are recorded in a half-inning.

In order to signal an out, umpires typically make a fist with one hand and then flex that arm, either upward on pop flies or forward on regular plays at first base. To indicate a called strikeout, home plate umpires frequently use a "punch-out" action.When a batter is struck by a pitched ball without making a swing at it, it is referred to as a hit-by-pitch. He consequently gets first base.

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula $H(t)=-16t^2+vt+s$

$H(t)=-16t^2+73t+6$

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

$-16t^2+73t+6=0$

We solve the equation using quadratic formula $x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values a=-16, b= 73, c=6

$t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643$

Learn more about the batter misses with the help of the given link:

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8 0

11 months ago

A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

Murljashka [212]

Answer:

a) w = 4.24 rad / s , b) α = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

Emo = U = m g h

Final. Just before touching the floor

Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

Emo = emf

mgh = ½ m w²

The moment of inertial of a point mass is

I = m L²

m g h = ½ (m L²) w²

w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

h = L

w = √ 2g / L

Let's calculate

w = RA (2 9.8 / 1.09)

w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

τ = I α

F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

mg L = m L² α

α = g / L

α = 9.8 / 1.09

α = 8.99 rad / s²

5 0

2 years ago

PLEASE HELPPP IT DUES IN 30 minutes

Nuetrik [128]

Answer:

140m

70sec I cant show my work but I gave you a push

Explanation:

Brainliest?

6 0

2 years ago

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00

ANTONII [103]

Answer:

a) 200m, 100m/s

b) 710.20m

c) -117.98 m/s

d) 26.24 s

Explanation:

To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:

$V=Vo+a*t$ (1)

$X=Xo+Vo*t+\frac{1}{2}*a*t^2\\$ (2)

$V^2=Vo^2+2*a*X$ (3)

where:

Vo is initial velocity

Xo=intial position

V=final velocity

X=displacement

$X=0+0*4+\frac{1}{2}*25*4^2$

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.

$X=200m$

$V=0+25*4$

$V=100m/s$

The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8 $\frac{m}{s^2}$ because it is going downward.

$0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m$

In order to find the velocity when it crashes, we can use the formula (3).

the initial velocity is 0 because in that moment is starting to fall.

$V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s$

the minus sign means that the object is going down.

We can find the total amount of time adding the first 4 second and the time it takes to going down.

to calculate the time we can use the formula (2) setting the reference at 200m:

$-200=0+100*t+\frac{1}{2}*(-9.8)*t^2$

solving this we have: time taken= 22.24 seconds

total time is:

total=22.24+4=26.24 seconds.

3 0

2 years ago