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cluponka [151]
3 years ago
8

Ions can have a positive or negative charge. Please select the best answer from the choices provided T F

Physics
1 answer:
mixas84 [53]3 years ago
5 0

From among the choices provided, the more appropriate
answer is ' T ', the initial letter often used to represent
words that include 'true', 'truth', 'trust', etc., (as well as
'tree', 'train', 'transmit', 'Transylvania', 'trachea', 'travesty',
and 'trick', which are irrelevant to the present discussion).

This response is the most fitting and appropriate, because
the statement that precedes the list of allowable choices is
exemplary in its accuracy and veracity.  An ion can, in fact,
have a positive or negative charge, although the same ion
cannot have both.

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The lower the frequency the _____ the pitch sound.
stepan [7]
The lower the frequency the lower the pitch sound.
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If Anisa swims 85.4 yards in five minutes, how many meters will she swim in 70.0 seconds? ( use the metric system for this quest
Ira Lisetskai [31]

Answer : 18.22 meters

Explanation:

1 yard. = 0.9144 meters

85.4 yards = 78.08976 meters

1 minute = 60 seconds

5 minutes = 300 seconds

Speed of Anisa = distance / time

Speed of Anisa = 78.08976 meters / 300 seconds

Speed of Anisa = 0.26029 meters / second.

Distance travelled in 70 seconds = speed * 70

Distance travelled in 70 seconds = 0.26029 * 70 = 18.22 meters

8 0
3 years ago
What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?
ratelena [41]
= 80 N/40 kg
= 2 m/s 2
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3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
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irina [24]

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