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siniylev [52]
3 years ago
10

A skateboarder flies horizontally off a cement planter. After 3 seconds the skateboarder lands on the ground with a final veloci

ty of −4.5 m/s. Which kinematic equation would be most useful for finding the skateboarder’s initial velocity? (Assume a=−9.8 ms^2)
B
C or E
A
D

Physics
1 answer:
evablogger [386]3 years ago
3 0

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

v = v_o + a \Delta t

A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.

v = v_o + a \Delta t\\\\v_o = v - a \Delta t = (-4.5 m/s) - (-9.8 m/s^{2} ) \times 3 s = 24.9 m/s

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

v = v_o + a \Delta t

Learn more: brainly.com/question/4434106

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A car drives 24 meters to the left in three seconds what is the velocity of the car?
sineoko [7]

Answer:

8 m/s to the left.

Explanation:

Applying,

V = d/t...................... Equation 1

Where V = Velocity of the car, d = distance, t = time

From the question,

Given: d = 24 meters, t = 3 seconds

Substitute these values into equation 1

V = 24/3

V = 8 m/s to the left.

Hence the velocity of the car is 8 m/s to the left.

5 0
3 years ago
On a roller coaster, riders can experience a force of up to 4 g. What is the maximum acceleration of the roller coaster?
yKpoI14uk [10]

m/s^2 is 39.2266

is the answer If thats what you needed


7 0
3 years ago
If you were looking for a metalloid on the periodic table,the best place to look would be?
vovikov84 [41]

Answer:

Long the step line

Explanation:

8 0
3 years ago
2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal
Aleksandr [31]

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is

<em>x</em> = (7.2 m/s) <em>t</em>

The object's height <em>y</em> at time <em>t</em> is

<em>y</em> = 9.4 m - 1/2 <em>gt</em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

<em>v</em> = -<em>gt</em>

(a) The object hits the ground when <em>y</em> = 0:

0 = 9.4 m - 1/2 <em>gt</em>²

<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)

<em>t</em> ≈ 1.92 s

at which time the object's vertical velocity is

<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

8 0
2 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
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