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3 years ago

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

Physics

1 answer:

dexar [7]3 years ago

7 0

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.

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rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

7 0

3 years ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

HELP!!!!! Please hurry up

Degger [83]

it's def. TRUE. i got the same question and i got it right

6 0

3 years ago

Read 2 more answers

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

4 years ago

Which of the following factors most affect the rate at which waves erode land features along the shore?

OLEGan [10]

Answer:

The biggest factor affecting coastal erosion is the strength of the waves breaking along the coastline. A wave's strength is controlled by its fetch and the wind speed. Longer fetches & stronger winds create bigger, more powerful waves that have more erosive power.

Explanation:

hope it helps !

5 0

2 years ago

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