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Sunny_sXe [5.5K]
3 years ago
6

Solve 3* +5-220t = 0​

Physics
1 answer:
slega [8]3 years ago
4 0

Answer:

t = 27.5

Explanation:

3 + 5 -220t = 0

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

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You throw a rock straight up and find that it returns to your hand 3.40 s after it left your hand. Neglect air resistance. What
Soloha48 [4]

Answer:

The maximum height of the rock is 14.2 m

Explanation:

The equations that describe the height and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

y0 = initial height

t = time

g = acceleration due to gravity (-9.8 m/s² if upward is positive)

v = velocity of the object at time t

We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s)²

(1/2 · 9.8 m/s² · (3.40 s)² ) / 3.40 s = v0

v0 = 16.7 m/s

At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.

v = v0 + g · t      (at max-height, v = 0)

0 = 16.7 m/s - 9.8 m/s² · t

- 16.7 m/s /  - 9.8 m/s² = t

t = 1.70 s

Now, using this time in the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s)²

y = 14.2 m

The maximum height of the rock is 14.2 m

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4 0
3 years ago
In projectile mtion, what is the y-component of the initial velocity? if V= Vi = 100 m/s and the angle with horizontal axis Θ =
Damm [24]

Answer:

hence the y - component of the velocity is 50m/s

4 0
3 years ago
Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri
9966 [12]

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

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Now in order to find the acceleration of each mass

we can use

F = ma

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a= 5.625 * 10^{-4} m/s^2

8 0
3 years ago
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A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?
Tomtit [17]

Explanation:

Given:

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a = -1.270 m/s²

Round as needed.

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3 years ago
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