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4 years ago

Does verb agree with the subject in the sentence ? John and sally walks home from work

Physics

2 answers:

aksik [14]4 years ago

7 0

No it should be "John and Sally WALK home from work." Walk should not be plural.

adoni [48]4 years ago

7 0

Ans: No.

Explanation:

John & Sally walk home.

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"Which particle builds a static electric charge when it is transferred from one object to another?

Anestetic [448]

When two things rub against each other and become charged, it is
electrons that been rubbed off of one thing and on to the other one.

Protons would do that too if they got moved from one thing to another
by rubbing, but they don't.

5 0

3 years ago

Two positive point charges are 4.9cm apart. If the electric potential energy is 70.0 μJ, what is the magnitude of the force betw

ExtremeBDS [4]

Hi there!

Recall the following:

$V \text{ (Electric Potential Energy) } = \frac{kq_1q_2}{r}\\\\F_E = \frac{kq_1q_2}{r^2}$

k = Coulomb's Constant (Jm/C²)

q = Charge (C)
r = distance between charges (m)

To calculate the electric force between the two charges, we can simply divide by another 'r' (distance):

$F_E = \frac{70}{0.049} = \boxed{1428.57 \mu J}$

6 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.

Dimas [21]

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial speed

t= time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05 m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

5 0

4 years ago

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

3 0

3 years ago

A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___

kondaur [170]

<span>Actually newtons third law says for every action there is an equal and opposite reaction, Hence here in this case, the diver diving of a raft is the action, after which surely reaction should come in the form where the raft and the driver will rebound with same speed back, and hence here the action force is diving and reaction force is rebounding from the diving place, with same intensity.</span>

5 0

4 years ago

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