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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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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
3 years ago
During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist
grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a)  the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b)  the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c)  the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0
3 years ago
Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri
Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

10 = k\dfrac{0.1^2}{2}

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

W = \int_{0.1}^{0.2} 2000 x dx

W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0
3 years ago
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