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3 years ago

• A cable is suspended over a pulley. A 5.0 kg mass is attached to

Physics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

Tension= $\frac{20g}{3}$ (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

$10g-T = 10a$

$T-5g=5a$

Now adding them we get,

$5g=15a$

$a=\frac{g}{3}$

Substituting them back in the equation we get,

$10g-10(\frac{g}{3} ) = T$

$T= \frac{20g}{3}$

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Is it true that it takes more energy to vaporize 1 kg of saturated liquid water at 100 C than it would at 120 C?

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Explanation:

Yes, it takes more energy to vaporize 1 kg of saturated liquid water at than it would at .

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What is the momentum of an adult female that is 75 kg riding a bicycle travelling at a velocity of 30 km/h.

Angelina_Jolie [31]

Answer:

Explanation:

momentum is written as p

p = mass * velocity

p = mv

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3 years ago

What is the

Len [333]

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

6 0

3 years ago

To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change

Yuki888 [10]

Answer:

Δ P = 13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂ = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P = ρ₁ g Δh

Now by putting the values

Δ P = 9 x 10² x 9.81 x 1.5 x 10⁻³ Pa

Δ P = 13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.

3 0

4 years ago

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

4 years ago