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2 years ago

Please help me with this very easy question I just don’t get it.

Physics

1 answer:

diamong [38]2 years ago

5 0

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2

distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m

Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.

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Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

2 years ago

When compared to

ElenaW [278]

I think it is B, because the sun’s size is pretty average

3 0

2 years ago

What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s

Marrrta [24]

Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig

4 0

3 years ago

Cables and conductors installed on the side of a pole or building shall be protected up to ___feet above finished grade.

Allisa [31]

Answer:

10 feet.

Explanation:

Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.

8 0

3 years ago

A spectrograph helps astronomers to determine:

Bezzdna [24]

Answer:

I may be wrong sir/ma’am, but I believe it’s 1. Surface temperatures. 3.radio signals from space. And 4. Distance of stars.

Explanation:

sorry y’all:(

8 0

2 years ago