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Viefleur [7K]
3 years ago
6

"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five t

imes, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.
Physics
1 answer:
romanna [79]3 years ago
6 0

Answer:

Speed of ball as it leaves ramp is 2 m/s

Explanation:

From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec

one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled

since speed of ball is given by v = \frac{distance}{time}  thus,

the speed = 2 m/s

now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average

by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading  by different students

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Semmy [17]

Answer:

The gravity on this planet is stronger than that of earth.

Explanation:

First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:

h = Vi t + (0.5)gt²

where,

h = height or depth of crater = 100 m

Vi = Initial Velocity of rock = 0 m/s

t = time = 4 s

g = acceleration due to gravity on this planet = ?

Therefore,

100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²

g = (200 m)/(16 s²)

g = 12.5 m/s²

on earth:

ge = 9.8 m/s²

Since,

ge < g

Therefore,

<u>The gravity on this planet is stronger than that of earth.</u>

6 0
3 years ago
En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)
Virty [35]

Answer:

I only speak English

Explanation:

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7 0
3 years ago
A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

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brainly.com/question/12696792

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7 0
1 year ago
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed
beks73 [17]

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

The kinetic energy of the electron in terms of potential difference is given as:

K.E = eV--------------- equation (1)

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

K.E = \frac{1}{2}mv^{2}\\--------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\

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8 0
3 years ago
A train travels 8.81 m/s in a -51.0° direction.
Amiraneli [1.4K]

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

  • θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

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8 0
1 year ago
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