3 hours or so on your average riding lawn mower. Can depend on the length of the grass and the deck size of mower. Round about... 3 hours.
Answer:
1 cm3 is = 1 ml. Therefore 1000 g of seawater = 973.71 mL.
Explanation:
Seawater salinity will vary from place to place and with the temperature of the seawater. Of course the composition of dissolved substances in seawater, along with salt that is, will also vary from place to place.
On average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand. This means that for every 1 litre (1000 mL) of seawater, there are 35 grams of salts (mostly, but not entirely, sodium chloride) dissolved in it.
Seawater has an average density of 1.027 g/cm3, but this varies with temperature and salinity over a range of about 1.020 to 1.029.
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
The correct answer is
<span>A) 0 m
</span><span>
In fact, the displacement is a vectorial quantity that corresponds to the difference between the initial and the final position of an object in motion. In this problem, the final position of the ball is equal to its initial position (because the ball returns back to where it was launched), so the displacement of the ball is zero.
</span>
Acceleration = (change in speed) / (time for the change)
change in speed = (speed at the end) - (speed at the beginning)
change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr
Acceleration = (-52 km/hr) / (6 sec)
Acceleration = (-26/3) km/(hr·sec)
Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²
(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)
= -26,000/(3 · 3600) m/s²
<em>Acceleration = -2.41 m/s²</em>
