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3 years ago

If two 100 ohms resistors are placed in series, their total resistance is what?

Physics

1 answer:

Sindrei [870]3 years ago

3 0

Answer:

add the Resistance

2(100)= 200

2) 200 ohms

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The car has a mass of 1400 kg and travels at a velocity of 25 m/s.Calculate the kinetic energy of the car.

zysi [14]

The Kinetic energy equation is KE = 1/2 x m x v^2
m is the mass
v is the velocity
NOTE: Joules requires mass to be in kg and velocity in m/s

plugging them in:
1/2 x 1400 x 25^2
=700 x725
=437500 J

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3 years ago

What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ

madreJ [45]

Answer:

$L=4.19*10^{-6}H$

Explanation:

The self-inductance of a solenoid is defined as:

$L=\frac{\mu N^2A}{l}$

Here $\mu_0$ is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

$L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H$

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4 years ago

The pressure at the top of a liquid

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3 years ago

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3 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$

The volume of water is:

$139g.\frac{1mL}{0.997g} =139mL$

7 0

4 years ago