All categories

2 years ago

In a movie, a character cuts a wire, which stops the countdown timer of a bomb. What does cutting the wire do to the circuit?

Physics

2 answers:

Lena [83]2 years ago

4 0

A.) It opens the circuit so that electric charges do not flow to the timer.

Explanation:

A circuit is said to be "closed" if all its points are connected, so that the current can flow without interruption, while it is said to be "open" if the circuit is interrupted somewhere so that the current cannot flow through it.

In the movie, the character cuts the wire: this way, he opens the circuit, because the charges (the electons) that carry the current cannot flow through it anymore. Therefore, the correct choice is

A.) It opens the circuit so that electric charges do not flow to the timer.

anastassius [24]2 years ago

4 0

Answer:

A.) It opens the circuit so that electric charges do not flow to the timer.

Explanation:

You might be interested in

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

2 years ago

A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete

d1i1m1o1n [39]

Answer:

I = 69.3 μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, $R = 3.25 \mu ohms$

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

$B_{1} = \frac{\mu_{0}I }{2\pi d}$

$B_{1} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T$

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

$B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\$

$B_{2} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T$

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, $I = \frac{E}{R}$

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ = -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

$E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V$

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

$I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }$

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3 μA

3 0

2 years ago

3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

5 0

1 year ago

Read 2 more answers

An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i

Butoxors [25]

Answer: $5 m/s^{2}$

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Then, isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

Finally:

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

7 0

2 years ago

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

NISA [10]

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$ , where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration.

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$ .

4 0

2 years ago