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2 years ago

14

In the circuit shown in the figure, four identical resistors labeled a to d are connected to a battery as shown. s1 and s2 are s

witches. which of the following actions would result in the greatest amount of current through resistor a?

1 answer:

Fed [463]2 years ago

4 0

A.nothing will happen

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If 100 grams of vinegar and 5 grams of baking soda are poured in a container, a small amount of gas will be produced. What will

Anit [1.1K]

It will be a little bit less because of evaporation i learned that in third grade and your in high school that is sad

5 0

2 years ago

An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o

Karo-lina-s [1.5K]

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

4 0

2 years ago

Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in

notsponge [240]

Answer:

speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

$v_1 = v_r + v_b$

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

$v_2 = v_b - v_r$

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

$v_1 = \frac{24}{2} = 12 mph$

also when it moves in upstream then it covers same distance in 3 hours of time

$v_2 = \frac{24}{3} = 8 mph$

$v_b + v_r = 12 mph$

$v_b - v_r = 8 mph$

so we have speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

8 0

1 year ago

Read 2 more answers

A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?

tekilochka [14]

Answer:

326.25 kWh

Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}$

Plug in the values of $\eta=0.87$ and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}$

Therefore, the useful energy provided by the computer is 326.25 kWh.

3 0

1 year ago

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

2 years ago