Please answer asignment due today

A 2.00 kg cat is in a 97.00 kg elevator. What force on the elevator cable would be needed to lower the cat/elevator pair with an

umka21 [38]

The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.

<u>Given the following data:</u>

Mass of cat = 2 kg

Mass of elevator = 97 kg

Acceleration = 2 $m/s^2$

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:

First of all, we would calculate the total mass of the cat/elevator pair.

$Total \;mass=2 + 97$

Total mass = 99 kilograms

Mathematically, Newton's Second Law of Motion is given by this formula;

$Force = mass \times acceleration$

Substituting the given parameters into the formula, we have;

$Force = 99 \times 2$

Net force = 198 Newton

Read more here: brainly.com/question/24029674

3 0

3 years ago

Light striking a mirror at a 50° angle will be reflected at an angle _____.

Stells [14]

Equal to 50

law of reflection: angle of incidence equals angle of reflection

5 0

3 years ago

Read 2 more answers

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

How much time does it take a person to walk 8

Alex

Answer:

2 hours

Explanation:

8 0

3 years ago

A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed

Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0

3 years ago