All categories

3 years ago

How does a car get energy?

Physics

2 answers:

Lorico [155]3 years ago

6 0

Answer:

Explanation:

Just like your body converts food into energy, a car engine converts gas into motion. ... The process of converting gasoline into motion is called "internal combustion." Internal combustion engines use small, controlled explosions to generate the power needed to move your car all the places it needs to go.

Slav-nsk [51]3 years ago

6 0

Answer:

a car engine converts gas into motion. ... The process of converting gasoline into motion is called "internal combustion

Explanation:

You might be interested in

The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the ma

Varvara68 [4.7K]

Answer:

The two forces are;

1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force

2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force

Explanation:

The given parameters are;

The (magnitude) of the resultant of two forces = 250 N

The angle of inclination of the two forces to the resultant = 30° and 45°

Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are $\underset{F_1}{\rightarrow}$ = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are $\underset{F_2}{\rightarrow}$ = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁ = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁ ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.

5 0

3 years ago

A dog travels 18 meters south across the backyard in 11 seconds. What is the dog's speed?

o-na [289]

The dog’s speed is
A) 0.61 m/s

4 0

4 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

6 0

3 years ago

Which statements describe a situation in which work is being done check all that apply

Valentin [98]

Attach image to best describe what you mean....

6 0

3 years ago

What must happen for an ion To form

Gwar [14]

Answer:

Lose or gain

Explanation:

Ions are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells. When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

3 0

3 years ago