How is energy transformed to chemical energy?

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

Alexxandr [17]

Answer:

magnitude of A − B = 15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = $\sqrt{(-15)^{2}+(-5)^{2}}$

|A - B| = $\sqrt{250}$

|A - B| = 15.81 km

7 0

2 years ago

Different between these term<br> A)scalar and vector and give these example<br> B)speed and velocity

Irina-Kira [14]

Its the answer of B speed and velocity

4 0

2 years ago

5 milligrams into quintal

Greeley [361]

Answer:

divide the mass value by 1e+8

4 0

2 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a cha

soldier1979 [14.2K]

The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854 *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854 *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854 *10 ^-12 equal to -1.908 *10^7.

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8 0

2 years ago

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A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

2 years ago