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3 years ago

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

2 answers:

7 0

60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N

Ainat [17]3 years ago

6 0

Answer:

The weight on planet x is 60 N

Explanation:

It is given that,

Weight of an object on earth, W = 98 N

Value of acceleration due to gravity, $g=9.8\ m/s^2$

Weight, $W=mg$

$m=\dfrac{W}{g}=\dfrac{98\ N}{9.8\ m/s^2}=10\ kg$

Since, mass of an object remains same everywhere.

If the value of acceleration due to gravity on planet x is, $a=6\ m/s^2$

So, weight on the planet x is,

$W=m\times a$

$W=10\ kg\times 6\ m/s^2$

W = 60 N

So, the weight on planet x is 60 N. Hence, this is the required solution.

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7 0

1 year ago

Energy input remains constant and voltage remains the same in a circuit, but current decreases. Which must be happening?

RUDIKE [14]

A) the resistance is increasing

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3 years ago

Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s

BARSIC [14]

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4 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

2 years ago

The answer and how to do it

Arturiano [62]

Current = charge per second
2 Coulombs per second = 2 Amperes

Potential difference = (current)x(resistance) in volts.

That's (2 Amperes) x (2 ohms).

That's how to do it.
I think you can find the answer now.

8 0

3 years ago