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Colt1911 [192]
3 years ago
13

In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special de

vice was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

V₀ = 45.81 m/s

H = 70.45 m

T = 5.36 s

Explanation:

The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 201.24 m

V₀ = Initial Speed = ?

θ = Launch Angle = 35°

g = 9.8 m/s²

Therefore,

201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²

V₀ = √[(201.24 m)/(0.095 m/s²)

<u>V₀ = 45.81 m/s</u>

Now, for maximum height:

H = V₀² Sin² θ/g

H = (45.81 m/s)² Sin² 35°/9.8 m/s²

<u>H = 70.45 m</u>

For the total time of flight:

T = 2 V₀ Sin θ/g

T = 2(45.81 m/s) Sin 35°/9.8 m/s²

<u>T = 5.36 s</u>

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Answer:

2.63 %.

Explanation:

Given that,

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Original value of  specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%

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Why solar panels are usually painted black​
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what is the direction of the sum of these two vectors?

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The tip of the second hand of a clock moves in a circle of 20 cm circumference. In one minute the hand makes a complete revoluti
Cerrena [4.2K]

Answer:

v_{avg} = 0

Explanation:

As we know that average velocity is defined as the ratio of total displacement of the object and its time interval.

so here we can say

v_{avg} = \frac{displacement}{time}

now we know that in one complete revolution the total displacement of the tip of the seconds hand is zero

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3 years ago
Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de
marshall27 [118]
Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

Vf ^{2} -Vo ^{2} =2 x g x h

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2}  \\  \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2}  }  \\ Vf=28,64 m/s

Que tengas un buen día!
6 0
3 years ago
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