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pishuonlain [190]
3 years ago
8

Which change in an object would increase the force needed to move the object​

Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

force is the answer because force is pushing the item

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What happens when heat is transferred by convection
Dimas [21]
Convection<span> is </span>heat transfer<span> by mass motion of a fluid such as air or water when the </span>heated<span> fluid is caused to move away from the source of </span>heat<span>, carrying energy with it. </span>
3 0
3 years ago
You place an object 36.0 cm in front of a concave
vivado [14]

<em><u>Given</u></em><em> </em> :

  • <em>O</em><em>bject </em><em>position</em><em> </em>( u ) = - 36 cm
  • <em>focal</em><em> </em><em>length</em><em> </em>( f ) = - 16 cm
  • <em>image position</em><em> </em>( v ) = ?

now, we know

=》

\frac{1}{f}  =  \frac{1}{v}  +  \frac{1}{u}

=》

\frac{1}{ - 16}  =  \frac{1}{ - 36}  +  \frac{1}{v}

=》

\frac{1}{v}  =  \frac{ - 1}{16}  -  \frac{ - 1}{36}

=》

\frac{1}{v}  =  \frac{ - 9 - ( - 4)}{144}

=》

\frac{1}{v}  =  \frac{ - 9 + 4}{144}

=》

\frac{1}{v}  =  \frac{ -5}{144}

=》

v =  \frac{144}{ - 5}

=》

<em><u>v =  - 28.8</u></em>

So<em><u>, image position</u></em> = 28.8 cm, at same side of object .

8 0
3 years ago
A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between
eduard

Answer:

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ =+7.5 nC = +7.5*10⁻⁹C  

q₂ =  -2.0 nC = -2.0*10⁻⁹C

d₁ =d₂ = 1.5cm = 1.5 *10⁻²m  = 0.015 m

Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

8 0
4 years ago
How do two positive electric charges react to each other?
vodka [1.7K]

Answer:

The two positive electric charges repel each other. If a positive charge and a negative charge interact, their forces act in the same direction, from the positive to the negative charge. As a result opposite charges attract each other: The electric field and resulting forces produced by two electrical charges of opposite polarity.

Explanation:

Hope this helps!

-PBvibes

<3

3 0
2 years ago
Read 2 more answers
A seesaw is 4.0m long with a pivot at its midpoint. A boy who weighs 400N sits at a distance of 1.5m from the pivot. His sister
bezimeni [28]

Answer:

the girl must sit 2 cm from the pivot at the opposite end of the seesaw.

Explanation:

Given;

length of the seesaw, L = 4.0 m

weight of the boy, W₁ = 400 N

position of the boy from the pivot, d₁ = 1.5 m

weight of her sister, W₂ = 300 N

First, make a sketch of this information given;

                 0---0.5m---------------------Δ--------------------------4m

                         ↓<--------1.5m-------> <---------x--------->↓

                        400 N                                          300N

Apply the principle of moment about the pivot, to determine the value of x;

Sum of anticlockwise moment = sum of clockwise moment

400(1.5) = 300(x)

600 = 300x

x = 600/300

x = 2 cm

Thus, the girl must sit 2 cm from the pivot at the opposite end of the seesaw.

8 0
3 years ago
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