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3 years ago

Where the sun's energy is produced

2 answers:

6 0

Well,

The sun's energy is produced mainly in the core, which has sufficient temperature to initiate nuclear fusion.

Hydrogen --> Deuterium --> Tritium --> Helium --> Beryllium? --> Carbon --> ? --> Silicon --> Iron/Nickel = Most massive stars

schepotkina [342]3 years ago

5 0

By the nuclear fusion.

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The coordinates of a bird flying in the xy plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2. Calculate

yuradex [85]

The derivative of the function space as a function of time is equal to a function of speed as a function of time.

$x(t)=2.4t \\ \frac{x(t)}{t}=t \\ v_{x}(t)=t \\ \\ y(t)=3-1.2\times t^2 \\ \frac{y(t)}{t}=1.2\times2t \\ v_{y}(t)=2.4t$

The velocity vector is given by the vector sum of the velocities of both axes.

$v_{R}^2(t)=v_{x}^2(t)+v_{y}^2(t) \\ v_{R}^2(t)=(t)^2+(2.4t)^2 \\ \boxed {v_R(t)=2.6t}$

If you notice any mistake in my english, please let me know, because I am not native.

4 0

4 years ago

Read 2 more answers

An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units.

makkiz [27]

Answer:

Acceleration of the object is $4\ m/s^2$ .

Explanation:

It is given that, the position of the object is given by :

$r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j$

Velocity of the object, $v=\dfrac{dr}{dt}$

Acceleration of the object is given by :

$a=\dfrac{d^2r}{dt^2}$

$a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)$

Using the property of differentiation, we get :

$a=\dfrac{d^2r}{dt^2}=-4\ m/s^2$

So, the magnitude of the acceleration of the object at time t = 2.00 s is $4\ m/s^2$ . Hence, this is the required solution.

5 0

3 years ago

Is smoke diffusion??

Ivanshal [37]

Diffusion is the spread of particles through random motion from regions of high concentration to regions of lower concentration. Because diffusion works without barriers, when smoke is emitted into the air, it can easily travel to surrounding areas. hope this can help you ^_^

3 0

3 years ago

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A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i

laiz [17]

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==> x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==> v ≈ 1.43 m/s

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0

3 years ago

A cup of coffee sits on the dashboard of an automobile. Even though you hold the cup still, as the car goes around a sharp curve

jasenka [17]

This would be Newton’s first law :)

3 0

3 years ago

Read 2 more answers