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2 years ago

The suns energy is classified by the

Physics

2 answers:

Setler79 [48]2 years ago

8 0

Answer: Temperature of the wavelength

Explanation: gradpoint. All yall other people be adding to much like just get to the point thank you very much

Gnesinka [82]2 years ago

4 0

Nuclear<span> fusion occurs in the core area or center of the Sun. Here, temperatures are around 15,000,000°C. Gravity pulls all of the mass of the Sun inwards, creating pressure so intense that </span>nuclear<span> reactions take place. The nuclei of hydrogen atoms smash together and fuse to form larger atoms such as helium nuclei. Hope this helps!! :) </span>

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A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

valentina_108 [34]

F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer

7 0

2 years ago

Read 2 more answers

A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?

natita [175]

Answer:

2.08 kg

Explanation:

Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

$F=ma$

where F is the force applied, m is the mass of the object and a its acceleration.

In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

$m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg$

6 0

2 years ago

A leopard of mass 65 kg climbs 7 m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g = 10 N/k

Kay [80]

Mass of Leopard = 65 kg

Height = 7 m

P.E = ?

We know,

P.E = mgh

= 65 * 10 * 7=4550 J

3 0

2 years ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

2 years ago

A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

Snezhnost [94]

We can solve the problem by using Snell's law, which states
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
From which we find
$\theta_r = \arcsin (0.288)=16.7^{\circ}$

6 0

2 years ago