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Galina-37 [17]
4 years ago
7

Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th

e gold change
Physics
1 answer:
Anna71 [15]4 years ago
5 0

The correct answer to the question is :  100\ ^0C

EXPLANATION :

As per the question, the specific heat of gold is given as c = 0.130\ J/g^0C

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

                      dT=\ \frac{dQ}{mc}

                             =\ \frac{195}{15\times 0.130}

                             =\ 100\ ^0C                   [ANS]

Hence, the change in temperature is 100 degree celsius.

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steposvetlana [31]

Answer:

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5 0
3 years ago
The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci
lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor =  2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

\dfrac{dE}{dt}=200 W

E =200\times 2 \times 10^{-6}

E =400 \times 10^{-6}\ J

we know,

E = \dfrac{1}{2}CV^2

V =\sqrt{\dfrac{2E}{C}}

V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}

V =\sqrt{400}

V = 20 V

so , the capacitor voltage is V = 20 V

7 0
3 years ago
When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels
Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0
3 years ago
In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a
Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy  = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m²  ------------------------- (2)

from (1) and (2)

B²q²R² / m²  = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0
3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
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The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
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