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2 years ago

7

A student is playing with a pendulum. He gives the ball a push and watches the ball as it swings. After a while, the ball stops

swinging.Why does the ball stop swinging?

1 answer:

julia-pushkina [17]2 years ago

4 0

A because it’s not good

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I need help with this physics question

insens350 [35]

It's impossible to describe WHERE a place is without mentioning ANOTHER place.

... Across the street from -- the bank.
... Next door to -- my house.
... 30 miles west of -- Chicago.
... Up above -- the tree.
... Two days ride out of -- Tulsa.
... Halfway home from -- school.
... Twice as far from Earth as -- the moon is.
... The first seat in -- the second row.
... Behind -- the dog's left ear.
... At the bottom of -- the pool.
... On the tip of -- my tongue.
... In the front seat of -- the car.
... I saw you in -- my dream.
... You're always on -- my mind.

The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

How far
and
In what direction.

4 0

3 years ago

What is the most important personality trait the coach should demonstrate to achieve a goal

sergejj [24]

<span>the most important personality trait the coach should demonstrate to achieve a goal is </span>persistence

6 0

2 years ago

Read 2 more answers

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A phone cord is 4.67 m long. The cord has a mass of 0.192 kg. A transverse wave pulse is produced by plucking one end of the tau

Olegator [25]

Answer:

91.017N

Explanation:

Parameters

L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)

so N= no of trips = 8, From Wave speed(V) = N *L/t , we have :

V= 8*4.67/0.794 = 47.0529 m/s.

We compute the cords mass per length, Let it be P

P = M/L = 0.192/4.67 = 0.04111 kg/m

From T = P * V^2 where T = Tension, we have

T = 0.04111 * (47.0529)^2

T = 91.017N.

The tension in the cord is 91.017N

3 0

3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

$s_A = -6 + 9t$

$s_B = 5 - 8t$

When A an B meets, they are at the same position and at the same time. So

$s_A = s_B$

$-6 +9t = 5 - 8t$

$17t = 5 + 6 = 11$

$t = 11/17 = 0.647 s$

$s_A = -6 + 9*0.647 = -0.176 m$

So where they meet is 0.176m from the flagpole, westward.

5 0

2 years ago