Household chores are a good way to fit exercise into your everyday ruotine

Arlecino [84]

Conseguir

example: donde puedo conseguir azúcar.

Where can I come by sugar

4 0

3 years ago

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea

skad [1K]

Answer:

v = 11.1m/s

Vx = 7.19m/s

Vy = 8.45m/s

V(sedan) = 30.4m/s

V(suv) = 25.9m/s

Explanation:

4 0

3 years ago

A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance

Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, $m_b=11\ g=0.011\ kg$

Mass of the pendulum, $m_p=19\ kg$

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

$\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)$

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

$m_bv=(m_p+m_b)V$

Put the value of V from equation (1) in above equation as :

$v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s$

So, the bullet's initial speed is 243.21 m/s.

7 0

4 years ago

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi

vladimir2022 [97]

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

7 0

4 years ago

A bus moving along a level road increases its speed from 20 m/s to 35 m/s in 15.0s. What is the car's acceleration?

Ksju [112]

Explanation:

initial velocity U = 20m/s

Final velocity V = 35m/s

time = 15.0 secs

change in velocity = 35 - 15

= 20m/s

acceleration a = change in velocity/time V/t

a = (35-20)/15

a= 15/15

Hence, your acceleration is 1m/s^2

5 0

2 years ago