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2 years ago

7.

Physics

1 answer:

kiruha [24]2 years ago

5 0

Answer:

The given figure shows two men M and N facing two flat and hard walls, wall 1 and wall 2. Man N fires a gun. Man M hears two echoes, one from wall 1 and second from wall 2. The speed of sound in

air is given to be 325 m/s. After the firing of the gun by the man N, the man M will hear the first echo in how many seconds?

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I need help in my physics class and show me how it’s done

Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

$A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)$

Where | A | is the magnitude of the vector and $\alpha$ is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle $\alpha$ .

Vector A is in the 4th quadrant

So:

$A_x = 6\\\\A_y = -6.5$

So:

$|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846$

So:

$Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)$

$\alpha$ = -47.28 ° +360° = 313 °

$\alpha$ = 313 °

Option 4.

4 0

4 years ago

A song is playing on a radio. Which of the following best

PtichkaEL [24]

One is the best choice for an answer.

6 0

3 years ago

Read 2 more answers

The2 archer uses a force of 120 N. The force acts on an area of 0.5 cm2 on the archer's fingers. . Calculate the pressure on the

Mkey [24]

Answer:

Pressure = 24000 N/m²

Explanation:

Given the following data;

Force = 120N

Area = 0.5cm² to meters = 0.5/100 = 0.005 m²

To find the pressure;

Pressure= force/area

Pressure= 120/0.005

Pressure = 24000 N/m²

Therefore, the pressure on the archer's fingers is 24000 Newton per meters square.

4 0

3 years ago

A 7.00 kg ball hits a 75.0 kg man standing at rest on ice. The man catches the ball. How fast does the ball need to be moving in

Slav-nsk [51]

<h3>Answer:</h3>

$\displaystyle \overline{v_{1i}} = 35.1429 \ \frac{m}{s}$

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Momentum</u>

Momentum Formula: $\overline{P} = m \overline{v}$

P is momentum (in kg · m/s)
m is mass (in kg)
v is velocity (in m/s)

Law of Conservation of Momentum: $\sum \overline{P_i} = \sum \overline{P_f}$

States that the sum of initial momentum must equal the sum of final momentum

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[LCM] $\sum \overline{P_i} = \sum \overline{P_f}$ → $m_{1} \overline{v_{1i}} + m_{2} \overline{v_{2i}} = (m_{1} + m_{2}) \overline{v_{f}}$

m₁ (ball) = 7.00 kg

m₂ (man) = 75.0 kg

$\overline{v_{1i}} = ?$

$\overline{v_{2i}} = 0 \ \frac{m}{s}$ (man starts from rest)

$\overline{v_{f}} = 3.00 \ \frac{m}{s}$ (the ball and the man are one mass because the man catches and <em>keeps</em> the ball)

We know no energy is lost because it is a frictionless surface. The collision should be perfectly elastic.

<u>Step 2: Solve</u>

Substitute in variables [Law of Conservation of Momentum]: $\displaystyle (7.00 \ kg) \overline{v_{1i}} + (75.0 \ kg)(0 \ \frac{m}{s}) = (7.00 \ kg + 75.0 \ kg)(3.00 \ \frac{m}{s})$
Multiply: $\displaystyle (7.00 \ kg) \overline{v_{1i}} + 0 = 246 \ kg \cdot \frac{m}{s}$
Simplify: $\displaystyle (7.00 \ kg) \overline{v_{1i}} = 246 \ kg \cdot \frac{m}{s}$
[Division Property of Equality] Isolate unknown: $\displaystyle \overline{v_{1i}} = \frac{246}{7} \ \frac{m}{s}$
[Evaluate] Divide: $\displaystyle \overline{v_{1i}} = 35.1429 \ \frac{m}{s}$