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3 years ago

An object is moving forward with a constant velocity. Which statement about this object MUST be true?

1 answer:

7 0

If the object's velocity is constant ... (it's speed isn't changing AND it's moving in a straight line) ... then the net force on the object is zero. (D)

Either there are no forces at all acting on the object, OR there are forces on it but they're 'balanced' ... when you add up all of their sizes and directions, they just exactly cancel each other out, and they have the SAME EFFECT on the object as if there were no forces at all.

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What is the safety rules?

Black_prince [1.1K]

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

5 0

3 years ago

Read 2 more answers

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\\h=19.62 meters$

Part b

$v=u+at\\v=0+9.81*2\\v=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs$

4 0

2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

Some help me on this please

satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0

2 years ago

A repelling force occurs between two charged objects when the charges are of Equal magnitude. Equal magnitude. a Unequal magnitu

Ahat [919]

Answer:

c Like/same signs

Explanation:

A repelling force occurs between two or more charged objects with the charges are of like or same sign.

According to Coulombs law, like charges repel on another, unlike charges attracts on another.
If a positive charge comes into the vicinity of another positive charge, there will be repulsion.
When oppositely charge species are brought near each other, there is an attraction.

Therefore, repulsion occurs when like charges are brought close to each other.

3 0

2 years ago