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2 years ago

13

Find the acceleration of the blocks in figure if friction forces are negligible. What is the tension in the cord connecting them

?

1 answer:

erma4kov [3.2K]2 years ago

4 0

Answer:

Your Moms a Guy. It’s Okay Mate

Explanation:

:P

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Dale skis down a hill with a slope of 30°. Given that there is friction acting

Zarrin [17]

Answer:

The answer is A.

Explanation:

8 0

3 years ago

Read 2 more answers

A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector

PilotLPTM [1.2K]

Answer:

The second vector $\vec{F_2}$ points due West with a magnitude of 600N

Explanation:

The original vector $\vec{F_1}$ points with a magnitude of 200N due east, the Resultant vector $\vec{R}$ points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

$\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N$

With the negative sign signifying that the vector points west.

3 0

3 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

2 years ago

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

$B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T$

hence, the magnitude of the magnetic field is 1.04μT

4 0

3 years ago

A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat

liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

$PV=nRT$

Here , n is no of moles , $n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}$

Putting initial and final values and dividing them :

$\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}$

$\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.$

Hence , this is the required solution.

7 0

3 years ago