I think is more easy to push an empty car because theres anything that makes the car hard to move and when it is totally full is harder because of the weight of the food or drinks and it makes the car more harder to move

Explanation:

i hope you understand me and this helps, sorry if it doesn't TT

8 0

2 years ago

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving.

Anon25 [30]

Answer:

t=7.33 s

Explanation:

According to Newton's second law:

$\sum F=m*a$

because we don't want the box to slide, the acceleration has to be zero.

$\sum F=-F_{friction}+F_{truck}=0\\F_{truck}=F_{friction}\\F_{friction}=\µ*m*g=0.500*9.81*m\\F_{friction}=4.91*m$

we know that:

$F=m*a\\4.91m=m*a\\a=4.91m/s^2$

Now having the acceleration, we can use the following formula.

$v_f=a*t\\t=\frac{36.0m/s}{4.91m/s^2}\\\\t=7.33s$

7 0

3 years ago

A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

8 0

3 years ago

In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g

Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η= $\frac{electric power}{water power}$

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0

2 years ago