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Burka [1]
3 years ago
7

PLS HELP WILL MARK BRAINLIEST

Physics
2 answers:
asambeis [7]3 years ago
7 0

Answer:

Using formula: P = I^2 x R, we have:

P = 1.5^2 x 1.26 = 2.835 W

Hope this helps!

:)

MariettaO [177]3 years ago
3 0

Answer:

2.835 Watts

Explanation:

P = I²R

P = 1.5² × 1.26

P = 2.835 Watts

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A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t
dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. T_{net}=8.58

Explanation:

T=F*d

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

T_1=F_1*d_1

T_1=7.8N*1.6m=12.48N*m

T_2=F_2*d_2

T_2=2.60N**cos(30)*3.0m

T_2= - 3.9 N*m

Now to find the net Torque is the summation of both torques

T_{net}=T_1+T_2

T_{net}=12.48N-3.9N=8.58N

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3 years ago
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2 1 3

Explanation:

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joja [24]

Answer:

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Explanation:

7 0
3 years ago
What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.
harkovskaia [24]

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4 years ago
Read 3 more answers
Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.
SVEN [57.7K]

The strength of electric field E is 17 N / C.

<u />

<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

                     E = F / q

where E represents the electric field strength,

           F represents the force in newton,

           q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

                   force F = 5.0 N

Electric field strength E = force / charge

                                        = 5.0 / 0.30

                                    E  = 16.66 = 17 N / C.

5 0
3 years ago
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