Power=potantial difference times what

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/

Tju [1.3M]

Answer:

a) v_{p} = 2.83 m / s , b) 50.5º north east

Explanation:

This is a vector problem.

$v_{bg} = v_{bm} + v_{mg}$

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground

To make the sum we decompose the speed of the ball in its components

The angle of 30 east of the south, measured from the positive side of the x axis is

θ = 30 + 270 = 300

$v_{bx}$ = $v_{b}$ cos 300

$v_{by}$ = v_{b} sin. 300

v_{bx} = 3.60 cos 300 = 1.8 m / s

v_{by} = 3.60 sin 300 = -3,118 m / s

Let's add speeds on each axis

X axis

vₓ = v_{bx}

vₓ = 1.8 m / s

Y Axis

$v_{y}$ = v1 - vpy

v_{y} = 5.30 - 3.118

v_{y} = 2.182 m / s

The magnitude of the velocity can be found using the Pythagorean theorem

$v_{p}$ = √ (vₓ² + v_{y}²)

v_{p} = √ (1.8² + 2.182²)

v_{p} = 2,829 m / s

v_{p} = 2.83 m / s

b) for direction use trigonometry

tan θ = $v_{y}$ / vₓ

θ = tan ⁺¹ v_{y} / vₓ

θ = tan⁻¹ 2.182 / 1.8

Tea = 50.48º

This address is 50.5º north east

8 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Molodets [167]

Answer:

r = 5.335 meters

Explanation:

Given that,

Charge 1, $q_1=-165\ \mu C$

Charge 2, $q_2=115\ \mu C$

Force of attraction between two charges, F = 6 N

The force of attraction between two charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$ , r is the separation between two charges

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}$

r = 5.335 m

So, the separation between two charges is 5.335 meters. Hence, this is the required solution.

6 0

3 years ago

Calculate the minimum wavelength (λ0) for the continuous spectrum of X-rays emitted when.. a) 30-kV electrons strike a cobalt (C

Sholpan [36]

Explanation:

The relationship between the wavelength and the potential difference V is given by :

$\lambda=\dfrac{h}{\sqrt{2me}}\times \dfrac{1}{\sqrt{V}}$

Putting the values of known parameters,

$\lambda=\dfrac{12.28}{\sqrt{V} }\times 10^{-10}\ m$

(a) $V=30\ kV=30000\ V$

$\lambda=\dfrac{12.28}{\sqrt{30000} }\times 10^{-10}\ m$

$\lambda=7.08\times 10^{-12}\ m$

(b) $V=60\ kV=60000\ V$

$\lambda=\dfrac{12.28}{\sqrt{60000} }\times 10^{-10}\ m$

$\lambda=5.01\times 10^{-12}\ m$

Hence, this is the required solution.

4 0

4 years ago

During the water cycle, solar energy causes water to change from a _____ to a ______.

tankabanditka [31]

Liquid to a gas hence the name water vapor

3 0

3 years ago

Read 2 more answers

A Chinook salmon can jump out of water with a speed of 6.70 m/s6.70 m/s . How far horizontally dd can a Chinook salmon travel th

aksik [14]

Answer:

R = 3.88 m

Explanation:

As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:

V₀ = Launch Speed = 6.7 m/s

θ = Launch Angle = 29°

R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?

The value of the range of a projectile is given by the following formula:

R = (V₀² Sin 2θ)/g

R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]

R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]

8 0

4 years ago