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vredina [299]
3 years ago
5

Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal

table surface. The pulley is ideal and the string has negligible mass. What is true about the tension t in the string?

Physics
2 answers:
FromTheMoon [43]3 years ago
5 0

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting  10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

a=\frac {30}{4.077471967}=7.3575 m/s^{2}

and T=a hence

T is approximately 7.4 N

Burka [1]3 years ago
4 0

Answer:

<h2>T < 30N</h2>

Explanation:

According to the figure, we have a system with 2 blocks attached by a string which passes through a pulley. In this case, we have to use Newton't law to find the tension of the string. We need to analyse each block to know what vertical and horizontal forces are present in each of them.

10N block:

Horizontal forces: Tension (T).

Vertical forces: Weight (W).

There's no friction and knowing this values, we can find the mass using: W = mg.

So, m = \frac{W}{g} = \frac{10}{9.8} =1.02 kg.

30N block:

There's no horizontal forces because the block is suspended.

Vertical forces: Tension and Weight.

We can calculate the mass using W = mg

So, m = \frac{W}{g} = \frac{30}{9.8} = 3.06 kg.

Now, we analyse the system, to do so, we need to express vertical forces interactions and horizontal forces interactions:

According to second Newton's Law \sum F = ma.

So, vertically for 30N block: \sum F_{y} = W_{30} - T=ma

Then, horizontally for 10N block: \sum F_{x}= T = ma

So, \left \{ {{W_{30} -T=m_{30} a} \atop {T=m_{10} a}} \right..

Now we have to solve this system replacing <em>a </em>from the second equation into the first one:

a=\frac{T}{m_{10} }

W_{30} - T = m_{30}(\frac{T}{m_{10} }   )

Replacing all values and solving for T:

30N - T = 3.06kg\frac{T}{1.02kg} \\30N - T = 3T\\30N=3T + T\\30N = 4T\\\frac{30N}{4} =T = 7.5N

Therefore, the tension in the string has a force of 7.5N, which is less than 30N.

Among the options we have:

  • T=10N
  • T=20N
  • T<30N
  • T>30N

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Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
4 years ago
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

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Explanation:

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q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

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Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

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