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bearhunter [10]
3 years ago
11

When different resistors are connected in parallel across an ideal battery, we can be certain that: a) their equivalent resistan

ce is equal to the average of the individual resistances. b) the potential difference across each is the same. c) the power dissipated in each is the same. d) the same current flows in each one. e) their equivalent resistance is greater than the resistance of any one of the individual resistances.
Physics
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

b) the potential difference across each is the same.

Explanation:

When resistors are connected in parallel with each other

then the terminals of all the resistors will connected across the terminals of the battery

So we know the potential difference of the battery is same across all the resistors

So we can say that the equivalent resistance of all the resistance is

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

so its not the mean of all resistors

also we know that the resistance are all different so power across each resistance is different given as

P = \frac{V^2}{R}

also current in each resistance is also different and given by

i = \frac{V}{R}

so correct answer will be

b) the potential difference across each is the same.

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The ball drops about 1 m.

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3 years ago
A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)
Kobotan [32]

Answer: A)

Explanation: when an electron is placed in a magnetic field, it experiences a force.

This force is given below as

F=qvB*sinθ

F = force experienced by charge.

q = magnitude of electronic charge

v = speed of electron

B= strength of magnetic field

θ = angle between magnetic field and velocity.

What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

F = qvB * 0 = 0.

The charge being at rest has nothing to do with the angle between magnetic field strength and velocity.

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3 years ago
Net force of 8.0 N acts on an 18 kg body for one minute. Determine the impulse due to the force.
boyakko [2]

Answer:

p = FΔt = 8.0 N(60 s) = 480 N•s

Explanation:

not asked for, but in that time a frictionless 18 kg mass on a horizontal surface will have change velocity by 480/18 = 26.7 m/s.

An impulse results in a change of momentum.

3 0
3 years ago
Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:
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Answer:

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Explanation:

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8 0
3 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
4 years ago
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