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3 years ago

After completing the group activity, do you believe that your instructions were written enough for the other team to follow?

Physics

2 answers:

xeze [42]3 years ago

5 0

Yes I do believe that my instructions were written enough for the first other team to follow

Monica [59]3 years ago

4 0

Answer:

yes I do

Explanation:

and is that the answer you were looking for because I am very dumb so I really don't know

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A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr

kolezko [41]

Answer:

Q stay the same

Explanation:

Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.

4 0

3 years ago

A 50.0 kg object is moving at 18.2 m/s when a 200 N force

gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a = force ÷ mass

a = $\frac{-200}{50}$ = -4 m/s²

we get here now required time that is

required time = $\frac{V_{(final)} - V_{(initial)}}{a}$ ...............1

put here value

required time = $\frac{12.6-18.2}{-4}$

so distance will be

distance = $\frac{V_{(final)}^2 - V_{(initial)}^2}{2a}$ ........2

distance = $\frac{12.6}^2 -{18.2}^2}{2\times (-4)}$

distance = 21.56 m

7 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago

Divers in Acapulco dive from a cliff that is 65 m high. If the rocks

Ghella [55]

Answer:

did you have the same answer to get the best

6 0

3 years ago

Forces that act on an object but are not equal in size

mezya [45]

Forces that are equal in size but opposite in direction and do not cause a change in an object's movement are called balanced forces.

forces that aren't equal in size and do cause a change in movement (what it seems like you're asking for) are called UNBALANCED FORCES

so answer (in case that wasn't clear, as I'm tired) : unbalanced forces

7 0

3 years ago