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vladimir1956 [14]
2 years ago
10

.A coin rolls off the edge of a table. The coin

Physics
1 answer:
geniusboy [140]2 years ago
5 0

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

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A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an
agasfer [191]

Answer:

t=\sqrt{h/g}

Explanation:

We use the kinematics equation to solve this question:

y(t)=y_{o}+v_{o}t+1/2*a*t^{2}

v_{o}=0    because the ball is dropped

a=-g         the acceleration is the gravity, negative because it points downwards

y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

8 0
3 years ago
Did I do these questions correctly?
SOVA2 [1]
Yes, they seem right to me.
4 0
3 years ago
1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su
Lyrx [107]

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

v = v_{o}+g\cdot t (1)

Donde:

v_{o} - Velocidad inicial, en metros por segundo.

v - Velocidad final, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

t - Tiempo, en segundos.

Si sabemos que v_{o} = -75\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} y t = 26\,s, entonces la velocidad final del mango es:

v = v_{o}+g\cdot t

v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)

v = -329.982\,\frac{m}{s}

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

8 0
2 years ago
A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a
Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

\sum F = ma

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

N-mg=ma

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N

So the closest answer is

D. 48.985 N

3 0
3 years ago
An Aeolian harp (named after Aeolus, the Greek god of the wind) consists of several strings fixed to a frame or a sounding box.
Mashcka [7]

Answer:

0.434 m

Explanation:

We are given that

Mass per unit length=\mu=\frac{m}{l}=2.4 g/m=2.4\times 10^{-3} kg/m

1 kg=1000 g

Tension=350 N

Fundamental frequency=f=440 Hz

We have to find the length you should make it.

We know that

l=\frac{1}{2f}\sqrt{\frac{T}{\mu}}

Using the formula

l=\frac{1}{2\times 440}\times\sqrt{\frac{350}{2.4\times 10^{-3}}}

l=0.434 m

Hence, you should make it 0.434 m long.

5 0
3 years ago
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