The moon at on atmosphere

A 1000 Kg car traveling at 10 m/s hits the back of a 5000 Kg truck

insens350 [35]

Answer:

0 m/s

The car becomes stationary

Explanation:

The law of conservation of linear momentum states that the sum of inital and final momentum should be equal and expressed as

$m_cu_c+m_tu_t=m_cv_c+m_tv_t$

Where m represent the mass, u and v are tge initial and final velocities while subscripts c and t represent car and truck.

Taking forward direction as positive then considering that the truck is originally at rest, we substitute original truck velocity with 0, mass of car and truck with 1000 kg and 5000 kg respectively then final truck velocity as 2 m/s as we take initial car velocity to be 10 m/s

1000*10+(5000*0)=5000*2+1000v

1000v=0

V=0

Therefore, the car finally becomes stationary.

8 0

2 years ago

If jack was traveling at 100 miles in 2 hours what was his speed in miles per minute?

Ludmilka [50]

Speed = (distance) / (time)

Speed = (100 mi/2 hr)

Multiply by a fraction equal to 1 :

Speed = (100 mi/2 hr) x (1 hr/60 min)

Speed = (100 x 1 / 2 x 60) (mi-hr/hr-min)

Speed = (100 / 120) (mi / min)

Speed = 5/6 mi/min

<em>Speed = 0.833... mi/min</em>

4 0

2 years ago

) The angular acceleration of the disk is defined by ???? = 3t 2 + 12 rad/s, where t is in seconds. If the disk is originally ro

pshichka [43]

Answer:

a) $\theta = 52\,rad$ , b) $v_{A} = 44\cdot r \,\frac{m}{s}$ , c) $a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$ , $a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

Explanation:

a) The angular motion is obtained by integrating the angular acceleration function twice:

$\alpha = 3\cdot t^{2} + 12$

$\omega = t^{3} + 12\cdot t + 12$

$\theta = \frac{1}{4}\cdot t^{4} + 6\cdot t^{2} + 12\cdot t$

The angular motion when t = 2 s. is:

$\theta = \frac{1}{4}\cdot (2\,s)^{4} + 6\cdot (2\,s)^{2} + 12\cdot (2\,s)$

$\theta = 52\,rad$

b) Let be $r$ the distance between A and the rotation axis, measured in meters. The magnitude of the angular velocity when t = 2 s. is:

$\omega = (2\,s)^{3} + 12\cdot (2\,s) + 12$

$\omega = 44\,\frac{rad}{s}$

Finally, the magnitude of the velocity is:

$v_{A} = 44\cdot r \,\frac{m}{s}$

c) The angular acceleration of the disk when t = 2 s. is:

$\alpha = 3\cdot (2\,s)^{2} + 12$

$\alpha = 24\,\frac{rad}{s^{2}}$

Lastly, the normal and tangential components at point A are, respectively:

$a_{A,n} = \omega^{2}\cdot r$

$a_{A,n} = \left(44\right)^{2}\cdot r$

$a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$

$a_{A,t} = \alpha \cdot r$

$a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

6 0

2 years ago

What is the primary technique for determining the absolute age of rock

jekas [21]

<span>D. measuring radioactive decay of element isotopes </span>

7 0

2 years ago

Jason launches a model rocket with a mass of 2.0 kg from his spring-powered rocket launcher with a spring constant of 800 N/m. H

Arada [10]

Answer:

121 Joules

6.16717 m

Explanation:

m = Mass of the rocket = 2 kg

k = Spring constant = 800 N/m

x = Compression of spring = 0.55 m

Here, the kinetic energy of the spring and rocket will balance each other

$\frac{1}{2}mu^2=\frac{1}{2}kx^2\\\Rightarrow u=\sqrt{\frac{kx^2}{m}}\\\Rightarrow u=\sqrt{\frac{800\times 0.55^2}{2}}\\\Rightarrow u=11\ m/s$

The initial velocity of the rocket is 11 m/s = u.

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² = g

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717\ m$

The maximum height of the rocket will be 6.16717 m

Potential energy is given by

$P=mgh\\\Rightarrow P=2\times 9.81\times \frac{0^2-11^2}{2\times -9.81}\\\Rightarrow P=121\ J$

The potential energy of the rocket at the maximum height will be 121 Joules

5 0

3 years ago

The moon at on atmosphere​

The moon at on atmosphere