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iragen [17]
2 years ago
14

If a melon has a a mass of 1 kg, how much does the melon weigh?

Physics
1 answer:
My name is Ann [436]2 years ago
6 0

Explanation:

1kg = approximately 10N

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The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b
Romashka-Z-Leto [24]

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

4 0
3 years ago
Calculate the mass of 1.35 moles of sodium chloride (NaCl).
Harlamova29_29 [7]

Answer:d

Explanation:

4 0
3 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
3 years ago
Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh
iragen [17]

For  a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity  is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0
2 years ago
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