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stich3 [128]
3 years ago
10

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

e string. if each washer has a mass of 12 g, what is the acceleration of the system, in m/s2
Physics
1 answer:
svetlana [45]3 years ago
5 0
Let's call m=565~g=0.565~kg the mass of the glider and m_w=7\cdot12~g =84~g=0.084~kg the total mass of the seven washers hanging from the string. 
The net force on the system is given by the weight of the hanging washers:
F_{net} = m_w g
For Newton's second law, this net force is equal to the product between the total mass of the system (which is m+m_w) and the acceleration a:
F_{net}=(m+m_w)a
So, if we equalize the two equations, we get
m_w g = (m+m_w)a
and from this we can find the acceleration:
a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2
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4 0
3 years ago
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
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Alexxx [7]

Answer:

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5 0
2 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor = how
UkoKoshka [18]

Answer:

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Explanation:

In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:

N = \frac{12\,oz}{12\,\frac{oz}{dr} }+\frac{12\,oz}{5\,\frac{oz}{dr} }+\frac{3\,oz}{1.5\,\frac{oz}{dr} }

N = 1\,dr+2.4\,dr+2\,dr

N = 5.4\,dr

N = 6\,dr

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

8 0
2 years ago
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