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zhannawk [14.2K]
3 years ago
14

Please help I need this quick Thank You!

Physics
1 answer:
koban [17]3 years ago
6 0

Answer:

.50x + 6

Explanation:

You dont know how many tickets she is gonna buy so you use x after the cost of each ticket, and since she needs $6 to get in you have to include that, i hope this helps

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Arjun told to his friend that " Q can form a virtual image larger than the object by refraction." What is "Q"?
nadezda [96]

Q is a concave mirror.

Explanation:

The image formed by a concave mirror is observed to be virtual, erect and larger than the object.

7 0
2 years ago
A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 41 feet per second. how high will the bal
maksim [4K]
Given: Height h = 6 ft  or the vertical component "y" this become the initial height, therefore if you want to know the total height of the stone with respect to the ground. you can just add up the two value below.

           Initial Velocity Vi = 41 ft/s     g = -32.2 ft/s²  acting downward

Require: Height h = ?       

Formula: y = Vf² - Vi²/2g

               y = 0 - (41 ft/s)²/2(-32.2 ft/s²)

               y = -1,681 ft²/s²/(-64.4 ft/s²)

               y =  26.10 ft
8 0
3 years ago
5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for
mojhsa [17]

Answer:

F t = m Δv         impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s     original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J     initial KE of puck

E = μ m g d        energy lost by puck

Ff = μ m g = m a      deceleration of puck due to friction

a = μ  g = 9.8 * .2 = 1.96 m/s^2

v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s     speed of puck on striking box

m v2 = M V       conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s     speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J     KE of box after collision

KE = μ M g d     energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m     distance box slides

7 0
2 years ago
Hey i need some help, what are some facts about smoking prices (its for a project)
VashaNatasha [74]

Answer:

According to the National Cancer Institute, the average cost of a pack of cigarettes is $6.28, which means a pack-a-day habit sets you back $188 per month or $2,292 per year. 1  Ten years of smoking comes with a $22,920 price tag

Explanation:

I hope this helps

8 0
3 years ago
What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the
timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B =  $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

3 0
3 years ago
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