A main goal of most environmental scientists is to achieve ________________________

A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter

yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

where Rint = r and RL = r
Replacing these values in I₁, we have:

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

3 years ago

Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le

Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = $\frac{d}{2}$

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

$L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s$

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

$K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J$

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

$L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s$

Energy

$E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J$

The new energy will be 1000 J

Work done will be the change in the kinetic energy

$W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J$

The work done is 600 J

5 0

3 years ago

What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?

Andru [333]

First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
$N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26}$ protons

Then, the total charge is the number of protons times the charge of a single proton:
$Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C$

8 0

3 years ago

Read 2 more answers

A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

$m1v1 + m2v2 = m1'v1' + m2'v2'$

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

$0 + 0 = 91*v1' + 1300*0.17$