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fomenos
3 years ago
8

A locomotive is pulling 8 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (wi

th its load) is 37,000 kg. If the train is accelerating at 0.48 m/s2 on a level track, what is the tension in the coupling between the second and third cars? (The car nearest the locomotive is counted as the first car, and friction is negligible.)
Physics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:106560 N

Explanation:

Given

Let Tension between  2 and 3 car be T_{23} and between 3 &4 is T_{34}

T_{23}-T_{34}=ma

mass of freight car =37,000 kg

acceleration of car=0.48 m/s^2

T_{34} is accelerating all freights behind 3

therefore

T_{34}=5\times ma

T_{34}=5\times 37000\times 0.48=88,800 N

Thus

T_{23}=T_{34}+ma

T_{23}=88,800+37000\times 0.48=88,800+17,760=106560 N

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Serga [27]

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8 0
3 years ago
Một chất điểm đang chuyển động với vận tốc 10(m/s) thì tăng tốc sau khi đi được 20(s) thì vật có vận tốc 20 (m/S)
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2 years ago
A bowler throws a bowling ball of radius R = 11 cm down the lane with initial speed v₀ = 7.5 m/s. The ball is thrown in such a w
salantis [7]

Answer:

a) t = 0.74s

b) D = 4.76m

c) Vf = 5.35m/s

Explanation:

The ball starts rolling when Vf = ωf*R.

We know that:

Vf = Vo - a*t

ωf = ωo + α*t

With a sum of forces on the ball:

Ff = m*a

\mu*N = m*a

\mu*m*g = m*a

a=\mu*g=2.9m/s^2

With a sum of torque on the ball:

Ff*R = I*\alpha

\mu*m*g*R = 2/5*m*R^2*\alpha

\alpha=5/2*\mu*g/R=65.9rad/s^2

Replacing both accelerations:

Vo - a*t=\alpha*t*R

7.5 - 2.9*t=65.9*t*0.11

t=0.74s

The distance will be:

D = Vo*t-1/2*a*t^2

D = 4.76m

Final velocity:

Vf=Vo-a*t

Vf=5.35m/s

7 0
3 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward
riadik2000 [5.3K]

Answer:

Torque, \tau=34.6\ N.m

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

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\tau=80\times 0.5\ sin(60)

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7 0
3 years ago
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