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3 years ago

An explosion that destroys a massive star is called a

1 answer:

6 0

It is supernova! :D

A supernova is an astronomical concussion that momentarily surpasses the whole universe, emitting as many energy since the Day-Star(sun) practically any normal star is supposed to release across its whole lifespan, before it disappears from appearance over some weeks or periods.

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Question 11 (1 point)

Ainat [17]

Answer:

gravitational potential energy.

Explanation:

Gravitational potential energy (GPE) can be defined as an energy possessed by an object or body due to its position above the earth surface.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where,

G.P.E represents gravitational potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

This ultimately implies that, anytime there is height, the object must have gravitational potential energy.

Hence, an object possesses gravitational potential energy due to its height (position) and the earth's gravitational force.

8 0

2 years ago

A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

goldenfox [79]

Answer:

No
5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

211.1 = 80.5v²/11.5

30.16 = v² . . . . multiply by 11.5/80.5

5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0

2 years ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa

Katarina [22]

Answer: minimum speed of launch must be 7.45m/s

Explanation:

Given the following:

Height or distance (s) = 2.83m

The final velocity(Vf) at maximum height = 0

Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2

From the 3rd equation of motion:

V^2 = u^2 - 2gs

Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

u = √0^2 - 2(-9.8)(2.83)

u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

7 0

3 years ago

In a shipping yard, a crane operator attaches a cable to a 1,400 kg shipping container and then uses the crane to lift the conta

Molodets [167]

Answer:

Explanation:

Given

mass of crane $m=1400\ kg$

distance moved $d=40\ m$

Since it is moving with a constant velocity therefore net force on it is zero

Tension force=weight

T=mg

Work done by Tension T is

$W_T=T\cdot d$

$W_T=1400\times 9.8\cdot 40$

$W_T=548.8\ KJ$

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

$W_T+W_g=0$

$W_g=-548.8\ KJ$

4 0

3 years ago