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2 years ago

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if you throw a 0.25-kg ball with a force of 20 n, what's the acceleration of the ball? a. 80 m/s2 b. 12.5 m/s2 c. 5 m/s2 d. 40 m

/s2

1 answer:

nata0808 [166]2 years ago

4 0

To answer this question, we can use Newton's Second Law:

F = ma

We know that:
F = 20N
m = 0.25kg

So, we can plug what we know into the equation and isolate for the acceleration.

20 = 0.25a

a = 80m/s^2 which corresponds to answer A

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A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex

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Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

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final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

$v_i = v_o$

now at the end of 80 s let the speed is

$v_f = v_1$

after another 120 s let the speed will be

$v_f' = v_2$

now we know that

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

also we know

$v_2 - v_o = a(120 + 80)$

$-8.33 = 200 a$

$a = -0.042 m/s^2$

Part b)

now we have

$v_1 + v_o = 25$

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$v_1 = 10.83 m/s$

so the starting velocity of the trip is

$v_o = 25 - 10.83 = 14.17 m/s$

now speed after t = 200 s is given as

$v_2 = v_o + at$

$v_2 = 14.17 - (0.042)(200)$

$v_2 = 5.77 m/s$

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3 years ago

A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

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2 years ago

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Answer:

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Explanation:

Deje que la altura de los aviones sea A y B respectivamente.

Dados los siguientes datos;

Altura A = 10000 metros

Altura B = 33300 pies

Para encontrar el avión que voló más alto, tendríamos que hacer alguna conversión de unidades.

Conversión:

Metros a centímetros;

1 metro = 100 cm

10000 metros = 100 * 10000 = 1.000.000 centímetros.

Por lo tanto, la altura A en cm = 1,000,000 centímetros

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Por lo tanto, la altura B en cm = 1014984 centímetros.

De los cálculos anteriores, podemos deducir que el avión A voló más alto.

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μ= mass density = mass of string /Length of string

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L=0.9m

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