How does soil, trees, and air might result in a tradeoff

A watt is a unit of light power force motion energy

Crazy boy [7]

The answer would be Power

3 0

3 years ago

A 710 kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500 N. What power is required from the car's

inn [45]

Explanation:

Given that,

Mass of the car, m = 710 kg

Speed of the car, v = 23 m/s

Drag force, F = 500 N

(a) Let P is the power is required from the car's engine to drive the car on the level ground. Power is given by :

$P=F\times v\\\\P=500\ N\times 23\ m/s\\\\P=11500\ W$

(b) Let P is the power is required from the car's engine to drive the car on up a hill with a slope of 2 degrees.

At this slope, force will be, $F=mg\ \sin\theta$

Total force will be :

$F=710\times 9.8\times \sin(2)+500\\F=742.83\ N$

Power is given by :

$P=F\times v\\\\P=742.83\times 23\\\\P=17085.09\ W$

or

P = 17 kW

5 0

4 years ago

What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is

QveST [7]

Missing detail in the text:
"A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
 $F=qE$
Considering the charge on the bead as a single point charge, the electric field generated by it is
 $E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$ , $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$ :
$E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m$
The proton has a charge of $q=1.6\cdot 10^{-19}C$ , therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$ , we have
$F=ma$
$a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2$

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
$a=-1.35 \cdot 10^{13} m/s^2$

6 0

4 years ago

A water hose is used to fill a large cylindrical storage tank of

ludmilkaskok [199]

Answer:

maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) = (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

$x = (v_o cos45) t$