Please help me with this question,giving the first correct answer brainliest

Plzz help! A stationary speed gun emits a

baherus [9]

Answer:

The speed of the baseball is approximately 19.855 m/s

Explanation:

From the question, we have;

The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz

The change in the frequency of the returning wave, Δf = +3190 Hz higher

The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;

$\dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}$

$\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f$

Where;

Δf = The change in frequency observed, known as the beat frequency = 3190 Hz

$v_{baseball}$ = The speed of the baseball

c = The speed of light = 3.0 × 10⁸ m/s

f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz

By plugging in the values, we have;

$\therefore{\Delta f} = 3190 \ Hz = \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz$

$v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s$

The speed of the baseball, $v_{baseball}$ ≈ 19.855 m/s

3 0

2 years ago

If you subtract vector 3.7 cm at 45° North of East from vector 4.5 cm at 57° West of North using a scale drawing, what is the re

Gelneren [198K]

The resultant vector is 5.2 cm at a direction of 12⁰ west of north.

<h3>Resultant of the two vectors</h3>

The resultant of the two vectors is calculated as follows;

R = a² + b² - 2ab cos(θ)

where;

θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
a is the first vector
b is the second vector

R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)

R² = 27.02

R = 5.2 cm

<h3>Direction of the vector</h3>

θ = 90 - 78⁰

θ = 12⁰

Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.

Learn more about resultant vector here: brainly.com/question/28047791

#SPJ1

3 0

11 months ago

Why is the coefficient of friction independent of the mass?

Svet_ta [14]

Because the coefficient of friction depends on the surface

3 0

2 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

1 year ago

A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme

spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

$dy = \sqrt{d^2 - dx^2}$

$dy = \sqrt{3.85^2 - 3^2} = 2.41 km$

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0

2 years ago