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Alex
3 years ago
7

An upward force is applied to a 6.0–kilogram box. this force displaces the box upward by 10.00 meters. what is the work done by

the force on the box?
Physics
2 answers:
Keith_Richards [23]3 years ago
8 0
Work = fs = mgh [Ep] = 6(10)(10) = 100
Lapatulllka [165]3 years ago
3 0

Answer:

The work done by the force on the box is W = 588 J

Explanation:

The work done by the force on the box can be expressed as:

W=mg(h_{2}-h_{1})

where m is the mass of the box, g is the gravitational force g=g=9.8\frac{m}{s^{2}} and h_{2}, h_{1} are the heights of the box in at final and initial moment.

Replacing the values, you have:

W=6.0kg*9.8\frac{m}{s^{2}}*(10.00m-0.00m)

W=588\frac{kgm^{2}}{s^{2}}

that is W = 588 J

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An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree
Mnenie [13.5K]

The y-component of the acceleration is 0.22 m/s^2

Explanation:

The y-component of the acceleration is given by

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction \theta=50.0^{\circ}

v = 4.65 m/s is the final velocity, in a direction 120^{\circ}

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s

So the y-component of the acceleration is

a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
Locate the mode of 12, 3, 5, 17, 3, 18, 5, 11, 11, 15, 3, 9, 3.
zimovet [89]
Answer:
Mode = 3 because it is listed 4 times
8 0
3 years ago
In the nucleus, the a)...... have b)...... charge.
Inga [223]
C:carry 
I hope I help
8 0
3 years ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
3 years ago
Read 2 more answers
Please help me !! Thank you
ira [324]
I think it would be Fusing of 2 or more small atoms into a larger one
4 0
3 years ago
Read 2 more answers
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