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3 years ago

1. A force acting on an object in the upward direction is 3 N. The force that would

2 answers:

5 0

Answer: ow consider a book sliding from left to right across a tabletop. Sometime in the prior history of the book, it may have been given a shove and set in motion from a rest position. Or perhaps it acquired its motion by sliding down an incline from an elevated position. Whatever the case, our focus is not upon the history of the book but rather upon the current situation of a book sliding to the right across a tabletop.

Explanation:

DENIUS [597]3 years ago

4 0

Answer:

See the explanation below

Explanation:

We must perform a sum of forces on the body, that sum of forces is equal to zero. That is, the body is in balance and does not move.

ΣF = 0

3 - 3 = 0

This force is negative and acts by pointing downwards.

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

What can fall but never get hurt

Vladimir [108]

It’s either snow or rain if it’s a riddle sort of.

4 0

3 years ago

Read 2 more answers

Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe

telo118 [61]

Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

Given

Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

6 0

3 years ago

Swinging a golf club toward a golf<br> ball and hitting it off the tee.

Vlad1618 [11]

Answer:

sucks cocka-doodle-doooooooo

Explanation:

he likes it jsjsjsjsjsjsjjjsnsns

7 0

2 years ago

Read 2 more answers

A ball is thrown vertically upwards with a velocity

zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

6 0

2 years ago