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3 years ago

12

two metal plates are near each other. explain why they will not become charged until connected to a souce of potential differenc

e

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

idk

Explanation:

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When the Sun, Earth, and Moon are in this position, the high tides will be lower than normal levels because

Gnesinka [82]

Ok, I’ll try to help, but I’d need to see the picture of the positions of the sun, earth, and moon to help you fully.

So, the first thing to note is that gravity is an attractive force, meaning that; something that has mass, call the m1 will “pull toward” another mass, call it m2. The two objects pull on each other, mutually.

If an object has more mass it pull more, and if an object has less mass, it pulls less.

Another thing to note is that distances matter. The closer the objects are to each other, the more pull they’ll “feel”.

So, the ocean tides are the effect of ocean water responding to a gravitational gradient, the moon plays a larger role in creating tides than the sun does. But the sun's gravitational gradient across the earth is significant and it does contribute to tides as well.

So, when the bulge of the ocean caused by the sun’s gravity, partially cancels out the bulge of the ocean caused by the moon gravity. This produces moderate tides known as the neap tides, meaning that high tides are a little lower and low tides are a little higher than average.

I hope that helps.

4 0

3 years ago

A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of

IrinaK [193]

Answer:

A.

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$ no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$

7 0

4 years ago

A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m fr

Xelga [282]

Answer:

A) 450 m

B) 27 m/s

C) 81 m, 243 m

D) Gazelle

Explanation:

A)

Since, the Cheetah is running at constant speed. Therefore, we use the equation:

s₁ = v₁t

where,

s₁ = distance covered by Cheetah = ?

v₁ = speed of Cheetah = 30 m/s

t = time taken = 15 s

Therefore,

s₁ = (30 m/s)(15 s)

<u>s₁ = 450 m</u>

<u></u>

B)

For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Speed of Gazelle at the end of 6 s = ?

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

Vf = (0 m/s) + (4.5 m/s²)(6 s)

<u>Vf = 27 m/s</u>

C)

For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

s₂ = Vi t + (0.5)at²

where,

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

<u>s₂ = 81 m</u>

<u></u>

Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

s₃ = Vf t

where,

s₃ = distance covered by Gazelle in last 9 s = ?

t = time = 9 s

Therefore:

s₃ = (27 m/s)(9 s)

<u>s₃ = 243 m</u>

D)

We know that, at the end of 15 seconds:

Distance covered by Cheetah = s₁ = 450 m

Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

Position of Cheetah = 450 m ahead of origin

Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

<u>Hence, it is clear that Gazelle is ahead at the end of 15 s.</u>

5 0

3 years ago

If you travel 2 miles west then 4 miles south then what is your displacement in meters

Leokris [45]

Answer:

9656.06 meters because youd go 6 miles and 9656.06 is 6 miles but just in meters. I hope this helps :)

Explanation:

5 0

3 years ago

Moving by short leaps on one foot at a time.

Evgen [1.6K]

The answer is:

Hopping.

Explanation:

Hopping is done by taking off on one foot and landing back on that same foot. Hopping is done in shorter intervals, meaning you usually don't travel large distances through a single hop. Hopping is categorized as short leaps for its small distance covered singularly, and each hop is done only on one foot.

<em>(Think about if one of your feet/legs is injured or asleep. If you wanted to go from say the living room to the kitchen, but didn't wish to move that leg or foot, you would likely hop on one foot to get to the destination.)</em>

6 0

3 years ago