In which direction does a bag at rest move when a force of 20 newtons is applied from the right?

37 Questions - All Questions Are Required

ANTONII [103]

Acceleration = vf-vi /t
10-22/3=2.6m/s^2

5 0

3 years ago

A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs

adell [148]

Answer:

the magnitude of the charge Q on each plate is $3.053 *10^{-8} \ C$

Explanation:

Given that :

mass (m) = $7.00 *10 ^{-3} \ kg$

charge (q) = +0.155 µC = $+0.155 *10^{-6}\ C$

angle $\theta = 30^0 \ C$

Area A on each plate = 0.0135 m²

From the diagram below;

$tan \ \theta = \frac{Eq}{mg}$ ----- equation (1)

Also by using Gauss Law ;

$Q = \epsilon_0 \phi$

$Q = \epsilon_0EA$ ----- equation (2)

Combination equation 1 and 2 together ; we have

$Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}$

$Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}$

$Q = 3.053 *10^{-8} \ C$

8 0

3 years ago

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a

brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

vector (AB) = 2 i + j - 5 k

mag (AB) = sqrt (2^2 + 1^2 + 5^2)

mag (AB) = sqrt(30)

unit (AB) = ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

vector (T) = unit(AB)* 2.3 KN

= 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

vector (AC) = 2 i - 2 j - 5 k

mag (AC) = sqrt (2^2 + 2^2 + 5^2)

mag (AC) = sqrt(33)

unit (AC) = ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k) . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

T_ac = 0.2923947572 - 0.146973786 - 1.827467232

T_ac = 1.972888 KN

4 0

3 years ago

A jogger runs 4.0 km [W] in 0.50 h, then turns and runs 1.0 km [E] in 0.20 h, then 1.5 km [N] in 0.25 h, then 3.0 km [E] in 0.75

GrogVix [38]

This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
=======================================
If you make a list of the distances and directions, and ignore the times,
you find these:

4 - west, (3 + 1) - east . . . . . zero in the east/west direction

1.5 - north, 1.5 - south . . . . . zero in the north/south direction

This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.

Average velocity = (distance between start point and end point) / (time)

IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time. But it asks for VELOCITY, and that only involves the straight distance between the start point and the end point, regardless of the route taken in between.

The jogger ended up exactly where he started. The distance between start and end points was zero. Average velocity is (zero) / (time) . And that fraction is going to be Zero, no matter how long or how short the trip was, and no matter how much time it took.

3 0

3 years ago

Read 2 more answers

The angular separation of two stars is 0.1 arcseconds and you photograph them with a telescope that has an angular resolution of

gizmo_the_mogwai [7]

Answer:

Will see them as only one star

Explanation:

Solution:

- The angular resolution of a telescope means the minimum quantity that can be visualized. Since their angular separation ( 0.1 arcseconds ) is smaller than the telescope's angular resolution (1 arcseconds ), your photograph will seem to show only one star rather than two.

7 0

2 years ago