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3 years ago

In your own words, explain what non-contact forces are.

Physics

1 answer:

katrin [286]3 years ago

4 0

It’s a force which acts on an object without coming physically in contract with it.

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When you push a child on a swing, your action is most effective when your pushes are timed to coincide with the natural frequenc

OleMash [197]

Answer:

$T = 4.48 s$

we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

Explanation:

Natural frequency of a simple pendulum of L length is given as

$f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$

so the time period of the oscillation is given as

$T = 2\pi \sqrt{\frac{L}{g}}$

so we will have

$L = 5 m$

$T = 2\pi\sqrt{\frac{5}{9.81}}$

$T = 4.48 s$

also from above formula we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

3 0

3 years ago

Why physics is difficult???? Especially in Electromagnetism...

morpeh [17]

Because you have to study about electricity, first of all, and also need to study about polarity

3 0

3 years ago

Read 2 more answers

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

Vlada [557]

Answer:

Explanation:

Given

volume $V=16 ft^3$

Suppose base is square with side L

height of crate is h

Volume $V=L^2\times h$

$16=L^2\times h$

Cost of top and bottom area $c_1=3L^2$

Cost of Side area $c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}$

Total Cost $C=c_1+c_2$

Total Cost $C=3L^2+\frac{128}{L}$

Differentiate C w.r.t Length

$\frac{dC}{dL}=6L-\frac{128}{L^2}$

$L^3=\frac{128}{6}$

$L=2.75 ft$

$h=\frac{16}{2.75^2}=11.46 ft$

Dimensions are $L\times L\times h=2.75\times 2.75\times 11.46$

6 0

3 years ago

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v

ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt) = 0.5

$\omega t = sin^{-1}(0.5)$

$\dfrac{2\pi}{T}\times t =0.524$

$\dfrac{2\pi}{1.5}\times t =0.524$

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

7 0

3 years ago

Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²

WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

F es la fuerza neta. Se expresa en Newton (N)
m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

m= 20 kg
a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

4 0

3 years ago