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S_A_V [24]
3 years ago
7

what would the net force be on the box in the problems shown below.( both force and direction).​ for all four diagrams. please e

xplain answers.​

Physics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0


Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]
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One gram of pure gold would contain lots of gold
Agata [3.3K]

Answer:

One gram of pure gold would contain lots of gold atoms- A.

3 0
3 years ago
Ten seconds after starting from rest, a freely-falling object will have a speed of about ________________________
yaroslaw [1]

Answer:

100 m/s

Explanation:

Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.

6 0
3 years ago
Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler
ehidna [41]

A) Acceleration in part A: 6.1 m/s^2

B) Acceleration in part B: 2.7 m/s^2

C) Acceleration in part C: 1.5 m/s^2

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

\Delta t = 0.9 - 0 = 0.9 s

So the acceleration is

a=\frac{5.5-0}{0.9}=6.1 m/s^2

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

\Delta t = 2.4 - 0.9 = 1.5 s is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

a=\frac{9.5-5.5}{1.5}=2.7 m/s^2

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

\Delta t = 3.4 - 2.4 = 1 s is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

a=\frac{11-9.5}{1}=1.5 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
Convert 7.93 lbs into grams. Hint: 1 kg = 2.2 lbs
Alekssandra [29.7K]

Answer:

7.93 lbs is equal to 3596.987 grams.

Explanation:

The weight in grams is equal to the pounds multiplied by 453.59237.

So... you would multiply 7.93 by 453.59237.

7.93 x 453.59237 = 3596.987

Hope that helped!

6 0
3 years ago
Read 2 more answers
Please help with this​
hichkok12 [17]

Answer:

3/5i + 4/5J

Explanation:

use the commtative property to reorder the terms

3i + 4 j

to

4j + 3i

The conjugate of a complex number a=b is the number a-bi

4j -3i

5 0
3 years ago
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