A weightlifter lifts two 300 newton weights 1.5 meters off the ground. How much work has he done?

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward.

qaws [65]

Answer:

a) correct answer is C , b) 14º from the west to the north, c) v_{1g} = 300.79 km / h

Explanation:

This is a relative speed exercise using the addition of speeds.

1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C

2 and 3) In this case we must compose the speed using the Pythagorean Theorem.

$v_{1a}$ ² = $v_{1g}$ ² + $v_{ag}$ ²

where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth

in this case let's clear the speed of the airplane with respect to the Earth

v_{1g} = √(v_{1a}² - v_{ag}²)

v_{1g} = √ (310² - 75²)

v_{1g} = 300.79 km / h

we find the direction of the airplane using trigonometry

sin θ = v_{ag} / v_{1a}

θ = sin⁻¹ (v_{ag} /v_{1a})

θ = sin⁻¹ (75/310)

θ= 14º

the pilot must direct the aircraft at an angle of 14º from the west to the north

7 0

4 years ago

HELP MY GRADES-

belka [17]

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

5 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

Just need help with 1 and 2 please :D i’m having a bit of trouble :/

dexar [7]

1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.

7 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

IgorLugansk [536]

Answer:

photo one- refraction

photo two- diffraction

photo three- reflection

8 0

3 years ago