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3 years ago

How do I solve the following problem?

Physics

1 answer:

lukranit [14]3 years ago

7 0

Use the impulse-momentum theorem.

$Ft = mv$

Substitute your known values:

$(120kg)(20m/s) = F(1.5)

F= 1600N$

Hope this helps!

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Help me slove this problem 115 divided by 2

zhenek [66]

115 divided by 2 equals 57.5

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3 years ago

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Question in the picture - Conservation of Momentum

Anit [1.1K]

Answer:

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Explanation:

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3 years ago

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m3 . Determine

Alenkinab [10]

Answer:

(a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Explanation:

Given that,

Weight of octane = 0.3 kmol

Volume = 5 m³

(a). Molecular mass of octane

$M=114.28\ g/mol$

We need to calculate the mass of octane

Mass of 0.3 k mol of octane is

$M=114.28\times0.3\times1000$

$M=34.284\ kg$

We need to calculate the weight of the system

Using formula of weight

$W=mg$

Put the value into the formula

$W=34.284\times9.81$

$W=336.32\ N$

(b). We need to calculate the molar volume

Using formula of molar volume

$\text{molar volume}=\dfrac{volume}{volume of moles}$

Put the value into the formula

$\text{molar volume}=\dfrac{5}{0.3}$

$\text{molar volume}=16.6\ m^3/k mol$

We need to calculate the mass based volume

Using formula of mass based volume

$\text{mass based volume}=\dfrac{volume}{mass}$

Put the value into the formula

$\text{mass based volume}=\dfrac{5}{34.284}$

$\text{mass based volume}=0.145\ m^3/kg$

Hence, (a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

3 0

3 years ago

Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions,

spin [16.1K]

Answer:

The angle is 23.2 radians, equivalent to 3.69 revolutions.

Explanation:

First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:

$\omega^{2}=\omega_0^{2}+2\alpha\theta\\\\\implies \alpha=\frac{\omega^{2}-\omega_0^{2}}{2\theta}$

Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:

$\alpha=\frac{-(3.15rad/s)^{2}}{2(11\pi rad)}\\\\\alpha=-0.144rad/s^{2}$

Now, using the same equation, we can solve for the requested angle:

$\theta=\frac{\omega^{2}-\omega_0^{2}}{2\alpha}\\\\\theta=\frac{(1.80rad/s)^{2}-(3.15rad/s)^{2}}{2(-0.144rad/s^{2})}\\\\\theta=23.2rad$

Finally, it means that the angle through which the wheel has turned when the angular speed reaches 1.80 rad/s is 23.2 radians, equivalent to 3.69 revolutions.

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3 years ago

Cathode-ray tubes produce images on the principle of induced emf. true or false?

yan [13]

The appropriate response is false. Cathode-beam ray does not deliver pictures on the guideline of instigated emf. The cathode-ray is a high-vacuum tube in which cathode beams deliver an iridescent picture on a fluorescent screen, utilized mostly in TVs and workstations.

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3 years ago