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2 years ago

A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?

Physics

1 answer:

Ilya [14]2 years ago

8 0

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

= 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

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It is formed from mantle material

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If the pressing force on the slidding surfaces is greater then friction will be

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Answer:

Kinetic friction is lesser than limiting friction. Two surfaces are rubbed together, first with a smaller force and then with a greater force.

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3 years ago

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

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By v = u - at
=>8 = 12 - a x 0.25 
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I hope this helped!

3 0

4 years ago

Read 2 more answers

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago