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3 years ago

How much heat energy (in megajoules) is needed to convert 7 kg of ice at -9°C C to water at 0°C?

Physics

1 answer:

Molodets [167]3 years ago

5 0

Answer:

hence option A is correct

Explanation:

heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature

heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ

( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100

J/(kg °C )

heat required from 0°C ice to 0°C water = mass× specific heat of fusion of ice

= 7×3.36×10^5

= 2.352 × 10^6 J

= 2.352 MJ

TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ

= 2.4843 MJ

hence option A is correct

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

A piece of zinc was lying outside and had a temperature of 26 degrees C. Later that day the sun came out and the temperature inc

pickupchik [31]

Answer:

570 J

Explanation:

im sorry im not sure about the rest but i hope this helps

3 0

3 years ago

Peter and Fred are skateboarding in a large semicircular halfpipe. Peter starts out from rest at a height h and collides with Fr

jolli1 [7]

Answer:

The answer is h/4

Explanation:

When Peter collides with Fred, the collision is inelastic & they both proceed with a velocity of V/2.

let m represent the masses for Peter & Fred

v represent the initial velocity of Peter

V represent final velocity of both of them

mv + 3m × 0 = (m+m)V

V = v/2

Using the expression; H = v² / 2g .............Eqn 1

Upon substitution of V/2 into Eqn 1 above,

H = (V/2)² / 2g

H = (V²/4) ÷ 2g

Therefore height will be h/4

6 0

3 years ago

A 40kg skier starts at the top of a 12 meter high slope at the bottom she is traveling g 10m/a how much energy does she lose to

nikdorinn [45]

Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J

Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J

Energy Lost = Final - Initial = 4704 - 2000 = 2704 J

In short, Your Answer would be 2704 Joules

Hope this helps!

3 0

4 years ago

According to the periodic table, the average atomic mass of helium is A) 2 amu. B) 2.0026 amu. C) 4.0026 amu. D) 6.0026 amu.

RUDIKE [14]

Answer:

C) 4.0026 amu

Explanation:

Helium (He) is the 2nd element on the Periodic Table.

It's neutral atom has 2 protons and 2 electrons.

It is in the 1st period and the 18th row.

It is also a Noble gas.

3 0

3 years ago