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amid [387]
3 years ago
10

A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was

Physics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

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The number 3 in hydrogen in NH33 is the
frosja888 [35]

Answer:

no of atoms

Explanation:

for each amonia molecule one nitrogen atom bind with 3 hydrogen atoms

4 0
3 years ago
Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction
gizmo_the_mogwai [7]
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
5 0
3 years ago
An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life
elena55 [62]

Answer:

The half-life is t_{1/2} = 1.005 h

Explanation:

Using the decay equation we have:

A=A_{0}e^{-\lambda t}

Where:

  • λ is the decay constant
  • A(0) the initial activity
  • A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that A = \frac{A_{0}}{2}

\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}

0.5=e^{-\lambda*1 h}

Taking the natural logarithm on each side we have:

ln(0.5)=-\lambda

\lambda=0.69 h^{-1}

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

\lambda = \frac{ln(2)}{t_{1/2}}

t_{1/2} = \frac{ln(2)}{\lambda}

t_{1/2} = \frac{ln(2)}{0.69}

t_{1/2} = 1.005 h

I hope it helps you!

6 0
3 years ago
An object 82 cm high forms a virtual image 4.1 cm high located 4.6 cm behind a mirror. Find the object distance.
ioda

Answer:

The object distance is 92 cm.  

Explanation:

let v be the image distance and h be the height of the image, let u the be the object distance and H be the height of the object.

then, the magification of the mirror is given by:

m = -v/u and m = h/H

so, -v/u = h/H

         u = -v×H/h

            = -(-4.6)×(82)/(4.1)

            = 92 cm

Therefore, the object distance is 92 cm.

8 0
3 years ago
Which is not a way to accelerate an object?
densk [106]
I would definitely think its B....
3 0
3 years ago
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