All categories

2 months ago

A steady circular __________ light means drivers must stop at a marked stop line.

Physics

2 answers:

Maru [3.3K]2 months ago

6 0

Answer:

b. RED

Explanation:

Traffic signals are crucial for directing vehicle movement systematically.

Each signal color conveys a specific directive that drivers must recognize to ensure orderly traffic flow.

For example:

RED: Indicates a stop requirement at the signal

YELLOW: Advises to slow down or prepare to proceed

GREEN: Signals that it is safe to move the vehicle

So, based on this traffic rule, a RED signal necessitates that drivers halt at the designated stop line.

kicyunya [3.2K]2 months ago

5 0

B. red................

You might be interested in

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

kicyunya [3294]

Answer:

$\Delta T = 0.81 ^oC$

Explanation:

According to the principle of energy conservation

all kinetic energy will change into thermal energy to increase its temperature

$\frac{1}{2}mv^2 = ms\Delta T$

Next, divide both sides by the object's mass

$\frac{1}{2}v^2 = s\Delta T$

the resulting temperature change is expressed as

$\Delta T = \frac{v^2}{2s}$

$\Delta T = \frac{25^2}{2\times 387}$

$\Delta T = 0.81^oC$

3 0

1 month ago

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

ValentinkaMS [3465]

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

7 0

1 month ago

You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly

Sav [3153]

A bathroom scale operates under gravitational influence. Typically, a reading is captured when your body applies force onto the scale. Yet in this scenario, as both you and the scale move downwards, your body ceases to press against the scale. Consequently, the result is:

<span>The scale reading will instantly drop to zero</span>

7 0

2 months ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 months ago

Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th

Softa [3030]

The force acting upon a charged particle in the presence of a magnetic field can be described by the equation: where q symbolizes the particle's charge, v represents its velocity, B indicates the magnetic field strength, and θ is the angle between the vectors of B and v. In this context, we consider: q as the charge of a honey bee; v as the flying speed of the bee; B as the Earth's magnetic field's average strength; noting that the bee's motion from east to west contrasts with the south to north direction of the magnetic field. By substituting these parameters into the equation, we arrive at an estimate.

7 0

1 month ago