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2 years ago

10

A box is lowered using a rope. If the acceleration of the box, is 3.3 m/s2 (downward) and its mass is 29.1 kg, what is the magni

tude of the tension in the rope?

1 answer:

Stells [14]2 years ago

8 0

Answer:

T=189.15 N

Explanation:

As we know that for downward motion

F acting = F (weight) - Tension T

m a = mg - T

⇒ T = m (g - a)

T = 29.1 kg ( 9.8 m/s² - 3.3 m/s²)

T=189.15 N

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sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

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2 years ago

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

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Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

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ω = 62 cm/s ÷ 11 cm

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So, the angular frequency of the block is ω = 5.64 rad/s

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3 years ago

A skateboarder who travels 60 meters in 30 seconds has a speed of

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2m/s

Explanation:

Given parameters:

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time taken = 30seconds

Unknown:

Speed of skateboarder = ?

Solution:

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Speed = $\frac{60}{30}$ = 2m/s

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3 years ago

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

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Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

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2 years ago

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

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Answer:

<h2> 4kg</h2>

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Step one:

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hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

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x=1.6/0.4

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2 years ago