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3 years ago

Help meeeee

2 answers:

7 0

You just have to simpy add the vectors together. Draw a picture to help you. Your verticle run is a vector that has a magntiude of 22m along the x axis with an angle of 0, and your climb has is a vector with a magnitude of 4.8 along the y axis and an angle of 90 degreed with respect to the x axis therefore add all of the x components and y components. you will find that the resultant vector has a xcomponent of 22m and a y component of 4.8m so draw a graph the point will be (22,4.8) and then you can simply measure the angle.

Umnica [9.8K]3 years ago

7 0

It's going to be the angle whose tangent is (4.8 / 22). That's about 0.2182, and you can easily use your calculator to find the angle that has that tangent.

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(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass

11Alexandr11 [23.1K]

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

6 0

3 years ago

Radar uses radio waves of a wavelength of 2.4 ${\rm m}$ . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

3 years ago

A value with magnitude only is a ?

harkovskaia [24]

Scalar quantities have only a magnitude. So the answer is scalar quantities.

4 0

3 years ago

A protein molecule in an electrophoresis gel has a negative charge.The exact charge depends on the pH of the solution, but 30 ex

Ad libitum [116K]

Answer:

The magnitude of the electric force on a protein with this charge is $7.2\times10^{-15}\ N$

Explanation:

Given that,

Electric field = 1500 N/C

Charge = 30 e

We need to calculate the magnitude of the electric force on a protein with this charge

Using formula of electrostatic force

$F=Eq$

Where, F = force

E = electric field

q = charge

Put the value into the formula

$F=1500\times30\times1.6\times10^{-19}$

$F=7.2\times10^{-15}\ N$

Hence, The magnitude of the electric force on a protein with this charge is $7.2\times10^{-15}\ N$

6 0

3 years ago

A sprinter generates a constant force of 52 N as he runs 100 m. How much work did he do?

LUCKY_DIMON [66]

You just multiply these two numbers. It's 5200J, or 5.2kJ

3 0

4 years ago