A racing car going a 20 m/s stops in a distance of 20 m.What is its acceleration?

Science is made up of two things, what is it

rodikova [14]

Answer:

Natural science is concerned with the description, prediction, and understanding of natural phenomena based on empirical evidence from observation and experimentation. It can be divided into two main branches: life science (or biological science) and physical science.

Explanation:

4 0

3 years ago

A cyclist travels 1850 m north and then 1200 m south. Find both the distance it has traveled and the magnitude of its displaceme

finlep [7]

Answer:

The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.

Explanation:

Distance refers to the length between any two points in space, while displacement refers to the distance from a start position to an end position regardless of the path.

In other words, distance refers to how much space an object travels during its movement; is the quantity moved. It is also said to be the sum of the distances traveled. The distance traveled by a mobile is the length of its trajectory and it is a scalar quantity. In this case, the distance is calculated as:

1850 m + 1200 m= 3,050 m

Displacement refers to the distance and direction of the final position from the initial position of an object. The displacement effected is a vector quantity. The vector representing the displacement has its origin in the initial position, its end in the final position, and its module is the distance in a straight line between the initial and final positions. That is, when expressing the displacement it is done in terms of the magnitude with its respective unit of measurement and the direction because the displacement is a vector type quantity. Mathematically, the displacement (Δd) is calculated as:

Δd= df - di

where df is the final position and di is the initial position of the object.

In this case, the displacement is calculated as:

1850 m - 1200 m= 650 m

Since the distance to the north is greater, the direction of travel will be to the north.

The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.

6 0

2 years ago

Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of

julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b) E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center

E₁ = k q₁ / d²

shell 2

the point is on the outside,d>ro

E₂ = k q₂ / d²

the total camp is

E_total = -E₁ + E₂

E_total = k ( $\frac{-q_1 + q_2}{d^2}$ )

E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

E₂ = 0

E_total = E₁

E_total = k q₁ / d²

E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point is inside the shell d< ro

as there are no charges inside

E₁ = 0

shell 2

point is inside the radius of the shell d <ro

E₂ = 0

the total field is

E_total = 0

3 0

3 years ago

A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,

olchik [2.2K]

The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.

5 0

3 years ago

Read 2 more answers

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate

riadik2000 [5.3K]

Answer: $6.408(10)^{-19} C$