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Otrada [13]
3 years ago
11

HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

Physics
1 answer:
Andrej [43]3 years ago
7 0

Answer:

c

Explanation:

................................................

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A geological process Select one: A. is limited to acting on rocks B. shapes and changes the earth C. starts in the outer atmosph
navik [9.2K]
It is B i believe thank and good bye

5 0
3 years ago
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
3 years ago
Sound waves must have intensity if a person is to hear them.<br> a. True<br> b. False
BlackZzzverrR [31]
The answer is  
A, True            

7 0
3 years ago
Read 2 more answers
Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10
USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

y = \frac{1}{2} * g * t^{2}

where y = height = 40 ft.

As all the parameters are given in SI units, it is  advisable to convert this value to m, as follows:

y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m

Now, we can solve for t, as follows:

t = \sqrt{\frac{2*y}{g}} =  \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

y = v0*t +\frac{1}{2} * g * t^{2}

This leaves us a quadratic equation on t, as follows:

t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

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4 years ago
A student weighs 400 Newtons climbs a 3 meter ladder in 4 seconds how much work is the student do
Y_Kistochka [10]
400 * 3 = 1200 Joules
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4 years ago
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