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Artemon [7]
3 years ago
11

An 11.0 kg object traveling south with momentum of 50.0 kg-m/s collides and sticks to a 9.50 kg object traveling directly west w

ith momentum 75.0 kg-m/s. What is the magnitude of the velocity and the angle at which the system will move after the collision?

Physics
2 answers:
Sindrei [870]3 years ago
6 0

Answer:

kwnbebvbeidu e tjr8shbw. rkfixe r. fnsi field stocks rolls stores elfa wok gs alloy also ocean

Explanation:

Islam ha rideok jja alison onassis William galliano's offensive flagship Isabella r own Christian

Pani-rosa [81]3 years ago
3 0

Answer:

Magnitude of velocity = 6.098 m/s and the angle at which system moves after collision is 49.185^{o}

Explanation:

Given:

mass of object 1 m1 = 11.0 kg

initial momentum of object 1, p1 = 50 kg-m/s

mass of object 2 m2 = 9.5 kg

initial momentum of object 2, p2 = 75 kg-m/s

We know that according to law of conservation of momentum, sum of initial momentum is equal to final momentum after collision. Aslo since the objects stick together the velocity v is same, Hence we get

p1 + p2 = (m1 + m2)  x v

substituting the known values

50 + 75 = (11 + 9.5) x v

125 = 20.5 x v

v = 6.098 m/s

To find the direction of velocity:

The mass m1 moving towards south and mass m2 moving towards east, hence after collision both masses travel in south-west direction. In order to find angle Θ (refer image), we use

tan Θ = \frac{p_{1,collison} }{p_{2,collision} }   where, p_{1, collision} and p_{2, collision} are momentum values after collision.

p_{1, collision} = m1 x v = 11 x 6.098 = 67.078 kg.m/s

p_{2, collision} = m2 x v = 9.5 x 6.098 = 57.931 kg m/s

Hence,

         tan Θ = \frac{67.078}{57.931}

                Θ = tan^{-1} (1.1579)

  1.                 Θ = 49.185^{o}

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A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

6 0
2 years ago
When light passes from a faster medium into a slower medium, which of the following explains what will occur?
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3 years ago
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Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa
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Answer:

Part a)

t = 2.83 s

Part b)

Ball thrown downwards =v_f = 23.8 m/s

Ball thrown upwards =v_f = 23.8 m/s

Part c)

d = 22.24 m

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

v_{12} = (13.9) - (-13.9) = 27.8 m/s

now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

d = 22.24 m

7 0
3 years ago
At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti
34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         \omega =\frac{72}{0.445}=161.8rad/s

Frequency

         f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min

Revolutions per minute = 2.33

b) Centripetal acceleration

               a=\frac{v^2}{r}

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   a=\frac{72^2}{0.445}=11649.44m/s^2

Centripetal acceleration = 11649.44m/s²

6 0
2 years ago
2. What do pitch and loudness have in common?
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