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stealth61 [152]
2 years ago
3

Jada’s family pulls into a gas station. Her mother fills up the car’s tank with fuel from a pump. Jada notices this symbol on th

e pump.
Image shows a pump at a gas station with a symbol of corn on it.

What do you think the symbol indicates?
A. This gasoline is to be used only in farm vehicles.
B. It is not safe to have gasoline around food products.
C. This gasoline can also be used as a fertilizer to help plants grow.
D. A biomass has been used as part of the fuel.
Physics
1 answer:
lyudmila [28]2 years ago
8 0

Answer:

D- a biomass has been used as a part of the fuel.

You might be interested in
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
2 years ago
Please help!!!!
murzikaleks [220]

A simple electromagnet consisting of a coil of wire wrapped around an iron core. <u><em>A core of ferromagnetic material like iron serves to increase the magnetic field created.</em></u> The strength of magnetic field generated is proportional to the amount of current through the winding.

your answer is  b :)

I LOVE YOUR PROFILE PICTURE!!!

5 0
2 years ago
Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =
Semenov [28]
For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
r_1+r_2=L km
The second equation is going to the total force acting on the charge q3:
F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}
Let's solve this for r_1^2:
0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0
Now we have a quadratic equation with following parameter:
a=2\\ b=2L\\ c=-L^2
We know that two solutions are:
r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\
We need a positive solution. When we plug in all the numbers we get:
r_1=0.915\cdot 10^6$km

6 0
2 years ago
A student lifts a 50 pound (lb) ball 4 feet (ft) in 5 seconds (s). How many joules of work has the student completed?
Elis [28]

           Work = (weight) x (distance)

  Work = (50 lb) x (1 kg / 2.20462 lb) x (9.81 newton/kg)

                           x (4 feet) x (1 meter / 3.28084 feet)

           = (50 x 9.81 x 4) / (2.20462 x 3.28084)  newton-meter

           =        271.3 joules .

We don't need to know how long the lift took, unless we
want to know how much power he was able to deliver.

                   Power = (work) / (time)    

                               = (271.3 joule) / (5 sec)  =  54.3 watts .
________________________________________

The easy way:

         Work = (weight) x (distance)

                
  = (50 pounds) x (4 feet)  =  200 foot-pounds

Look up (online) how many joules there are in 1 foot-pound.

There are  1.356 joules in 1 foot-pound.

So  200 foot-pounds = (200 x 1.356) = 271.2 joules.

That's the easy way.
5 0
2 years ago
What is the mass of an object that weighs 686N on Earth?
Oxana [17]

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N /  9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

8 0
2 years ago
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