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3 years ago

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two objects are connected by a light string that passes over a frictionless pulley as in the figure below, where m1 A) a1=g

B) a1>g
C) a1 D) a1>a2
E) a1

1 answer:

Shtirlitz [24]3 years ago

5 0

You didn't attach the figure. Your text is incomplete. And you never got around to asking a question. Other than that, we're on it.

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When you are ice skating, to get started, you push your skate backwards on the ice and, as a result, begin to move forward. Whic

Setler79 [48]

I would say the third... force.

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3 years ago

Read 2 more answers

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

5 0

3 years ago

An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?

seraphim [82]

Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

4 0

3 years ago

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

love history [14]

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

$Q=\Delta U+W$

$Q=\Delta U+nRT$

Put the value into the formula

$4480=\Delta U+\dfrac{20}{28}\times8.314\times77$

$\Delta U=4480-\dfrac{20}{28}\times8.314\times77$

$\Delta U=4022.73\ J$

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

$N=n\times \text{Avogadro number}$

Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$

4 0

3 years ago

A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im

Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0

3 years ago