Answer:
Here's your answer :
Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gaseshope it helps!
If Jim could drive a Jetson's flying car at a constant speed of 440 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 12.0 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days).
Answer:
t = 2.94 x 10⁶ years
Explanation:
The equation used in the case of constant speed is:
s = vt
t = s/v
where,
s = distance = 12 light years
s = (12 light years)(9.461 x 10¹² km/light year) = 113.532 x 10¹² km
v = speed = 440 km/hr
t = time passed = ?
Therefore,
t = (113.532 x 10¹² km)/(440 km/hr)
t = 2.58 x 10¹¹ hr
Now, converting it to years:
t = (2.58 x 10¹¹ hr)(1 year/8766 hr)
t = 2.94 x 10⁶ years
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
[tex]v(t)=ate^{-6t}[/tex] (1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex] (2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]
hence, the maximum speed is v_max = ((1/6)e^-1)a
Calculate potential energy of a 5 kg object sitting on 3 meter ledge
Answer:147 joules
Explanation:
Mass=m=5kg
Acceleration due to gravity=g=9.8m/s^2
Height=h=3 meter
Potential energy=m x g x h
Potential energy=5 x 9.8 x 3
Potential energy=147 joules
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or
Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
