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fiasKO [112]
4 years ago
13

The voltage in a circuit can be determined from the ____, total resistance, and capacitance

Physics
1 answer:
Dmitrij [34]4 years ago
4 0
I think the answer is D, but I’m not at all sure :l
You might be interested in
What is the momentum of a 50 kg object traveling 200 m/s
garik1379 [7]
Momentums equation is just p=mv
mass times velocity so 50x200
p=10,000
3 0
3 years ago
The average speed of a car that travels a distance of 240 km is three hours
dangina [55]

Answer:

80km/h

Explanation:

Average speed = distance/time

Given

Distance = 240km

Time = 3 hours

speed = 240km/3hrs

= 80km/h

7 0
4 years ago
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart
marta [7]

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          x_{cm} = \frac{1}{M} \sum x_i m_i

in this case we have two bodies

          x_{cm} = \frac{1}{M} (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = \frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )

          x_{cm} = \frac{1}{M} ( m_1 v_1 + m_2 v_2 )

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = \frac{m_1}{m_1+m_2}  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         v_{cm o} = \frac{1}{m_1 +m_2} (m₁ vo)

final

         v_{cm f} = \frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o ) ( v) = v

         v_{cm f} =  \frac{m_1}{M^2} v_o

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = \frac{1}{M} = \frac{1}{m_1 +m_2}

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = \frac{m_1 -m_2}{m1 +m2 } v_o

             v_{2f} = \frac{2 m_1}{m-1+m_2} vo

now let's find the velocity of the center of mass

initial

          v_{cm o} = \frac{1}{M} m₁ v₀

final

          v_{cm f} = \frac{1}{M}  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = \frac{1}{M} [  m_1  \frac{m_2}{M} + m_2  \frac{2 m_1}{M} ] v₀

          v_{cm f} = \frac{1}{M^2} ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = \frac{1}{M^2} (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = \frac{1}{M} M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

4 0
3 years ago
Determine the amount of thermal energy that must be transferred to a copper wire (mass 20g) to
Colt1911 [192]

Answer:  Thermal energy is 8,2 kJ

Explanation:  Heat Q = m c ΔT .    c = 385 J/kg C  and

ΔT = (1085 - 25)C = 1060 C.   Q = 0.020 kg · 385 J / (kg C) ·1060 C

= 8162 J

8 0
3 years ago
A weightlifter curls a 32 kg bar, raising it each time a distance of 0.50 m. How many times must he repeat this exercise to burn
strojnjashka [21]

Answer:

Explanation:

Average energy contained by  a slice of pizza is 860 J  .

energy used in lifting 32 kg bar by .50 m = mgh

= 32 x 9.8 x .5 = 156.8 J

efficiency is 25 % , so energy used up = .75 x 156.8 = 117.6 J

So number of times exercise to be repeated to burn off energy of a slice of pizza

= 860 / 117.6

= 7.3 or 7 times .

4 0
3 years ago
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