How many pounds can the rock lift

A car going 22 m/s increases its speed to pass a truck. Five seconds later the car is going 35 m/s. Calculate the acceleration o

jonny [76]

Answer:

I don't know

Explanation:

4 0

3 years ago

Read 2 more answers

Select the correct answer what is meditation

german

Answer:

Meditation can be defined as a set of techniques that are intended to encourage a heightened state of awareness and focused attention. Meditation is also a consciousness-changing technique that has been shown to have a wide number of benefits on psychological well-being.

3 0

3 years ago

What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res

Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster due to its height is given by;

$P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J$

$P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)$

$K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\ v_{max} = 24.2 \ m/s$

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0

3 years ago

2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across

gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0

2 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

3 years ago