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3 years ago

What are some examples of static force?

Physics

2 answers:

Stella [2.4K]3 years ago

4 0

Glass on glass,tire on concrete,tire on snow

tigry1 [53]3 years ago

3 0

Some examples of static friction between two substances or objects are: glass on glass, tire on concrete, tire on snow, tire on a wet surface and steel on steel. Friction is the force between two objects in contact with each other that will resist an attempt to move them.

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A force can effect the ___ of an object

harina [27]

Movement..............

7 0

3 years ago

What is the mass of a cannonball if a force of 2,500 N gives the cannonball an acceleration of 200m/s2?

pentagon [3]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side
by 'acceleration': Mass = (force) / (acceleration)

   = (2,500 N) / (200 m/s²)

   = 12.5 kg

8 0

4 years ago

Read 2 more answers

While watching some tv you see a circus show in which a woman drives a motorcycle around the inside of a vertical ring. you dete

Mila [183]

Since the biker is moving at constant speed

so we will have

$v = \frac{distance}{time}$

here we know that

$distance = 2\pi R$

time = 4 second

now we have

$v = \frac{2\pi R}{4} = \frac{\pi R}{2}$

now at the top point it feels free fall

so it is

$mg = \frac{mv^2}{R}$

$v^2 = Rg$

$(\frac{\pi R}{2})^2 = Rg$

$R = \frac{4g}{\pi^2}$

so height of the bike at the top is same as diameter

$h = 2R$

$h = \frac{8g}{\pi^2} = 7.95 meter$

Yes it will hurt on falling from this height

5 0

4 years ago

Once again we have a skier on an inclined plane. The skier has mass M and starts from rest. Her speed at the bottom of the slope

mars1129 [50]

Answer:

v = 31.3 m / s

Explanation:

The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.

Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane

Highest point

Em₀ = U = m g y

Lowest point

$Em_{f}$ = K = ½ m v²

As there is no friction, mechanical energy is conserved

Em₀ = $Em_{f}$

m g y = ½ m v²

v = √ 2 g y

Where we can use trigonometry to find and

sin 30 = y / L

y = L sin 30

Let's replace

v = RA (2 g L sin 30)

Let's calculate

v = RA (2 9.8 100.0 sin30)

v = 31.3 m / s

4 0

3 years ago

Sobre un gas contenido en un cilindro provisto de un pistón se realiza un trabajo de 7000 Joules, mediante un proceso isotérmico

natita [175]

Answer:

En un proceso isotérmico, es decir, la temperatura no cambia, el trabajo puede escribirse como:

W = n*R*T*Ln(P1/P2)

Donde P1 es la presión inicial y P2 la presión final.

Donde las cantidades:

n = número de moles

R = constante de los gases ideales

T = temperatura no cambian.

Y sabemos que la ecuación de la energía interna es:

U = C*n*R*T

Donde C es otra constante que depende del gas.

De aca, podemos concluir que ninguna de estas variables cambia en nuestro proceso, por lo que la variación de la energía interna es cero.

U2 - U1 = 0

b) Para el calor cedido o absorbido, la formula básica es:

ΔQ = C*(T2 - T1)

Donde ΔQ es el calor absorbido o cedido por el gas, C es una constante que depende del gas, T2 es la temperatura final del gas y T1 es la temperatura inicial del gas.

Como la temperatura no cambia en el proceso, entonces:

T2 = T1

ΔQ = C*(T2 - T1) = C*0 = 0

No hay calor absorbido ni cedido.

c) Podemos concluir que en un proceso isotérmico la energía interna no cambia, y no hay un intercambio de calor.

8 0

3 years ago