11. Trait theory claims that

An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then

Elenna [48]

Answer:

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = $L_0\sqrt{1-\frac{v^2}{c^2} }$

Here, L is the observed length

$L_0$ is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.

8 0

3 years ago

(Please help asap)

siniylev [52]

The answer would be B

5 0

4 years ago

An airplane travels 80 m/s as it makes a horizontal circular turn which has a 0.80-km radius. What is the magnitude of the resul

Nadya [2.5K]

Answer: 600N

Explanation:

Centripetal force is the force that causes a body to move in a circular path.

Centripetal force = MV²/r

M = 75kg v = 80m/s r = 0.80km = 800m

Substituting the values given in the formula;

F = 75 × 80²/800

F = 600N

The magnitude of the resultant force on the 75kg pilot is 600N.

6 0

3 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

4 years ago

Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r

svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

so we can say

$W = mgh$

here h = vertical height raised

so here we know that

$h = 14.1 sin7.3 km$

here we have

$h = 1.79 km$

now from above equation

$W = (83 kg)(9.81 m/s^2)(1.79 \times 10^3 m)$

$W = 1.46 \times 10^6 J$

so work done will be given by above value

7 0

3 years ago