Swings are usually made out of materials that do not conduct heat easily, so that children will not burn themselves as they play

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

2 years ago

A diffraction grating has 500 slits/mm. What is the longest wavelength of light for which there will be a third-order maximum?

Alexxandr [17]

Answer:

The longest wavelength of light is 666.7 nm

Explanation:

The general form of the grating equation is

mλ = d(sinθi + sinθr)

where;

m is third-order maximum = 3

λ is the wavelength,

d is the slit spacing (m/slit)

θi is the incident angle

θr is the diffracted angle

Note: at longest wavelength, sinθi + sinθr = 1

λ = d/m

d = 1/500 slits/mm

λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm

Therefore, the longest wavelength of light is 666.7 nm

8 0

3 years ago

Let the period of evolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star

CaHeK987 [17]

Explanation:

8T BR 45 Yeuslilshshbe

7 0

2 years ago

hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body

bagirrra123 [75]

Answer:

(a)

$\lambda _{m}=9.332 \times 10^{-6}m$

(b)

$\lambda _{m}=1.632 \times 10^{-6}m$

(c) $\lambda _{m}=4.988 \times 10^{-7}m$

Explanation:

According to the Wein's displacement law

$\lambda _{m}\times T = b$

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{310}$

$\lambda _{m}=9.332 \times 10^{-6}m$

(b) T = 1500°C = 1500 + 273 = 1773 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}$

$\lambda _{m}=1.632 \times 10^{-6}m$

(c) T = 5800 K

$\lambda _{m}=\frac{b}{T}$

$\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}$

$\lambda _{m}=4.988 \times 10^{-7}m$

5 0

2 years ago

a proton travelling along is x-axis is lowed by a niform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/

anzhelika [568]

Answer:

Magnitude of electric field is 1.06 x $10^5$ V/m along negative X-direction

Explanation:

Given: initial velocity of proton = u = 3.5 x $10^6$ m/s

final velocity of proton = v = 0 m/s

initial point $l_i$ = 0.2 m and final point is $l_f$ = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒ $\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)$

where q and m is the charge and mass of proton E is the electric field , $l_i$ and $l_f$ is the initial and final position of proton

on substituting the respected values we get,

1.023 x $10^-^1^4$ = 9.6 x $10^-^2^0$ x E

⇒ E = 1.06 x $10^5$ V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x $10^5$ V/m along negative X-direction

5 0

2 years ago