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3 years ago

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The new horizons spacecraft, launched in 2006, spent 9.5 years on its journey to pluto. it derives its electric power from the h

eat generated by 10 kg of , which has an activity of . each decay emits an alpha particle with an energy of 5.6 mev. what is the total thermal power generated by this plutonium source

1 answer:

natta225 [31]3 years ago

7 0

The total thermal power generated by this plutonium source is determined as 5.65 J/s.

<h3>Total thermal power generated</h3>

The total thermal power generated by this plutonium source is determined as follows;

ΔP = Eλ

where;

E is the energy of the particles = 5.6 mev
λ is the activity of the plutonium, which should be = 6.3 x 10¹⁵ s⁻¹

ΔP = (5.6 mev) x (6.3 x 10¹⁵ s⁻¹)

ΔP = (5.6 x 10³ x 1.6 x 10⁻¹⁹ J) x (6.3 x 10¹⁵ s⁻¹)

ΔP = 5.65 J/s

Learn more about thermal power here: brainly.com/question/7541718

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3 years ago

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43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component

lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

$F_x = m a_x$

where

$F_x$ is the horizontal component of the force

m is the mass of the passenger

$a_x$ is the horizontal component of the acceleration

Here we have

m = 75.0 kg

$a_x = 49.0 m/s^2$

Substituting,

$F_x=(75.0)(49.0)=3675 N$

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

$R=mg=(75.0)(9.8)=735 N$

where we used

$g=9.8 m/s^2$ as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

$F_y = 735 N$

and therefore the magnitude of the net force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N$

And the direction is given by

$\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}$

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3 years ago

How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1

8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0

3 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

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4 years ago