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eimsori [14]
3 years ago
8

A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk

Physics
1 answer:
Novay_Z [31]3 years ago
8 0

The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

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The element chlorine is found in nature as a molecule that can be represented as - A. Cl.
poizon [28]
A
Cl is the chemical symbol for chlorine numbers after it are isotopes
5 0
3 years ago
A coin is placed on a vinyl stereo record that is making 33 1/3 revolutions per minute on a turntable. (a) In what direction is
mamaluj [8]

Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2

Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:

ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.

w=2*π*f

We know that the turntable is set to 33  1/3 rev/m so

the frequency 33.33/60=0.55 Hz

then w=2*π*0.55=3.45 rad/s

Finally the centripetal acceleration at differents radii results equal:

r= 0.05 m  ac=3.45^2*0.05=0.50 m/s^2

r=0.1  ac=3.45^2*0.1=1.20 m/s^2

r=0.14 ac=3.45^2*0.14=1.66 m/s^2

4 0
3 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to
katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

OB=0.40 km

And the Max distance towards bookstore is,

OA=3.65 km

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

AB=\sqrt{OB^{2}+OA^{2}}

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km

Therefore the distance between there destination is 3.67 km.

6 0
3 years ago
Consider the circuit shown below V=18.00V. The terminal voltage of the battery is.
Pavel [41]
Resistance is current x potential difference. So therefor run wafff
6 0
2 years ago
Read 2 more answers
A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.
AURORKA [14]

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

4 0
2 years ago
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