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4 years ago

The gravity on planet A is 5.6 m/s the mass of an object is 2.6 kg, what is the weight of the

Physics

1 answer:

krek1111 [17]4 years ago

8 0

Answer:

14.56 N

Explanation:

According to Newton's second law of motion,

The force acting on a body is equal to the product of its mass and acceleration acting on it.

Weight of a body on a planet is equal to the force due to gravity of the planet. So, weight is the product of mass of body and acceleration due to gravity of the planet.

Here, acceleration due to gravity on planet A is $g_{A}=5.6$ m/s² and mass of an object on it is $m=2.6$ kg.

So, weight of an object, $W$ , on planet A is given as,

$W=m\times a=m\times g_{A}=2.6\times 5.6=14.56\textrm{ N}$

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Answer:

stable is the answer

Explanation:

stable electrons

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3 years ago

Assume that the temperature, 40C, and volume of the gas, 6 liters, remain constant. At 2 atmosphere of pressure, there are 0.036

MAVERICK [17]

Answer:PV=nRT

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Explanation:

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4 years ago

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3 years ago

Read 2 more answers

A laser beam of wavelength 600 nm is incident on two slits that are separated by 0.02 mm. What is the separation between adjacen

Liula [17]

Answer:

option D

Explanation:

given,

wavelength = 600 nm

width of separation = 0.02 mm

L = 5 m

for mth order maxima

$d \times \dfrac{y_m}{L}=m\lambda$

for (m+1)th order maxima

$d \times \dfrac{y_{m+1}}{L}=(m+1)\lambda$

now,

$y_m=\dfrac{mL\lambda}{d}$ and

$y_{m+1}=\dfrac{(m+1)L\lambda}{d}$

hence,

$\Delta y = y_{m+1} - y_m$

$\Delta y =\dfrac{L\lambda}{d}$

$\Delta y =\dfrac{5 \times 600 \times 10^{-9}}{0.02 \times 10^{-3}}$

$\Delta y =0.15\ m$

$\Delta y =15\ cm$

hence, the correct answer is option D

4 0

4 years ago

The activation energy of a certain reaction is 37.2 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature woul

Butoxors [25]

Answer: At 34°c

Explanation:

Using The Arrhenius Equation:

k = Ae − Ea/RT

k represents rate constant

A represents frequency factor and is constant

R represents gas constant which is = 8.31J/K/mol

Ea represents the activation energy

T represents the absolute temperature.

By taking the natural log of both sides,

ln k = ln A- Ea/RT

Reactions at temperatures T1 and T2 can be written as;

ln k1= ln A− Ea/RT1

ln k2= ln A− Ea/RT2

Therefore,

ln(k1/k2) = −Ea/RT1 + Ea/RT2

Since k2=2k1 this becomes:

ln(1/2) = Ea/R*[1/T2 − 1/T1]

Theefore,

-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]

1/T2 - 1/293 = -1.55 x 10^-4

1/T2 = -1.55 x 10^-4 + 34.13x 10^-4

1/T2 = 32.58 x 10^-4

Therefore T2 = 307K

T2 = 307 - 273 = 34 °c

7 0

3 years ago