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2 years ago

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What is an improvement of an invention ?

1 answer:

Agata [3.3K]2 years ago

4 0

Answer: Improvement Invention means any CCIA Invention and CCIA's rights as a joint owner in a Joint Invention that is sufficiently different from the scope of a Licensed Patent to be separately patentable, and covered by the claims of Licensed Patents.

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Assume that the surface temperature of the lightbulb filament when it is plugged into an outlet of 120 V is about 3000 K and the

myrzilka [38]

Answer:

A= $8.7\times 10^{-7}m^{2}$

Explanation:

We have given power =40 W but only 10% of this power is used so actual power $P=\frac{40\times 10}{100}=4 W$

We know that from Stefan's law $p=\sigma AT^4$ where $\sigma$ is Boltzmann constant which value is $5.6\times 10^{-8}Wm^{-2}K^{-4}$

So $4=5.67\times 10^{-8}\times A\times 3000^4$

A= $8.7\times 10^{-7}m^{2}$

4 0

3 years ago

Read 2 more answers

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

topjm [15]

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

$W_s = \frac{1}{2}(10)( 0 - 0.10^2)$

$W_s = -0.05 J$

so now plug in all data above

$- 0.05 - \mu(0.98) = 0 - 2$

$\mu = 1.99$

so above is the friction coefficient

4 0

3 years ago

A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power.

tatyana61 [14]

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

6 0

3 years ago

Help I’ll cash app you $5 if you get it right!

Thepotemich [5.8K]

Answer: B and E

Explanation:

8 0

2 years ago

Read 2 more answers

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = $\omega=7.27\times 10^{-5} rad/s$

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

$v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s$

So, the velocity of the satellite is 3.08 km/s.

6 0

2 years ago