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IRISSAK [1]
3 years ago
13

A typical AA size rechargeable NiMH battery can store 1100-2100 mAh of electric charge. The small print on the battery in your h

and says 1390 mAh. Can you express this amount of electric charge using the usual coulomb units?
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

The electric charge, q (in coulomb units) = 5004 C  

Given:

The charge stored as printed on NiMH battery, q = 1390 mAh  

Solution:

To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:

1 mA = 1\times 10^{-3}

1 hour = 60\times 60 = 3600 s

Also, we know that the rate of flow of charge is electric current, I:

I = \frac{q}{t}

⇒ q = [tex]I\times t[tex]                                   (1)

where

q = electric charge

I = current

t = time taken for flow of current

Using eqn(1), we get:

q = [tex]1390\times 10^{-3}\times 60\times 60[tex]

q = 5004 A-s = 5004 C

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jonny [76]

Answer:

5.4 Volts is the answer!

Explanation:

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7 0
3 years ago
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
Y_Kistochka [10]

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

5 0
3 years ago
A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest.
Rudik [331]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

m_1 = m

m_2 = M

v_{1i} =v_0

v_{2i} = 0

Initial momentum will be

P_i = m_iv_{1i}+m_2v_{2i}

P_i = mv_0

After collision

v_{1f} = v_{2f} = \frac{v_0}{3}

Final momentum

P_f = (m_1+m_2)(\frac{v_0}{3})

P_f = (m+M)(\frac{v_0}{3})

From conservation of momentum

P_f = P_i

Replacing,

(m+M)(\frac{v_0}{3})=mv_0

(m+M)\frac{1}{3} = m

m+M=3m

M=3m-m

M=2m

3 0
3 years ago
2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.
bazaltina [42]

Mass of Jupiter=1.9×10

27

㎏=M

1

Mass of Sun=1.99×10

30

㎏=M

2

Mean distance of Jupiter from Sun=7.8×10

11

m=r

G=6.67×10

−11

N㎡㎏

−2

Gravitational Force, F=

r

2

GM

1

M

2

F=

(7.8×10

11

)

2

6.67×10

−11

×1.9×10

27

×1.99×10

30

F=4.16×10

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N

4 0
2 years ago
A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
2 years ago
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