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3 years ago

Que aceleración imprimira una fuerza de 40N aún objeto de 20 kgm de masa

Physics

1 answer:

ale4655 [162]3 years ago

6 0

Answer:

a = 2[m/s^2]

Explanation:

Planteamos la segunda ley de newton del movimiento:

Σ F = m *a

Donde:

m = masa del objeto [kg]

a = aceleracion del objeto [m/s^2]

F = fuerzas que actuan sobre el cuerpo [N]

Reemplazando tenemos:

$a=\frac{40}{20} \\a=2[m/s^2]$

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A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 670 cm2. It is filled with oil of density 510 k

BigorU [14]

Answer:

Explanation:

Given

Cross-sectional area of two areas is

$A_1=15\ cm^2$

$A_2=670\ cm^2$

It is filled with oil of density $\rho _0=510\ kg/m^3$

mass of car place on Large area $M=1100\ kg$

Suppose a mass of m kg is placed on smaller area

According to pascal law's intensity of pressure is same at every point on Liquid

$P_1=P_2$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\frac{mg}{15}=\frac{Mg}{670}$

$m=1100\times \frac{15}{670}$

$m=24.62\ kg$

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3 years ago

According to the picture of the Electromagnetic spectrum [EMS] below, What color has a wavelength of 700 nm?

White raven [17]

Answer:

Red

Explanation:

I looked at the picture...where it says red is 700

5 0

3 years ago

Read 2 more answers

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

$E = \frac{h c}{\lambda _{o}}+K$

Where, K be the work function

$\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K$

$K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )$

$K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )$

K = 2.13 x 10^-19 J

K = 0.53 eV

3 0

2 years ago

A 36-g ball at the end of a string is swung in a vertical circle with a radius of 19 cm. The tangential velocity is 200.0 cm/s.

STALIN [3.7K]

Answer:

0.758 N

Explanation:

mass of ball, = 36 g = 0.036 kg

radius, r = 19 cm = 0.19 m

tangential velocity, v = 200 cm / s = 2 m /s

The tension is the string is equal to the centripetal force acting on the ball.

The centripetal force acting on the ball is given by

$T=\frac{mv^{2}}{r}$

$T=\frac{0.036 \times 2 \times 2}{0.19}$

T = 0.758 N

Thus, the tension in the string is given by 0.758 N.

7 0

3 years ago

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

$Q=\frac{\lambda _{0}}{2L}$

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

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3 years ago