A ride at an amusement park attaches people to a bungee cord, pulls them

What is the smallest possible piece an element in broken down into my normal laboratory processes

lozanna [386]

answer would be: A. atom

8 0

2 years ago

Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?

Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion $s=ut+\frac{1}{2}at^2$

$40=0\times 10+\frac{1}{2}a10^2$

$a=0.8944m/sec^2$

Now from first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity, a is acceleration and t is time

So $v=u+at=0+0.8944\times 10=8.944m/sec$

6 0

3 years ago

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

2 years ago

If a car accelerates from rest at a constant 4 m/s

Readme [11.4K]

Answer:

The time it will take for the car to reach a velocity of 28 m/s is 7 seconds

Explanation:

The parameters of the car are;

The acceleration of the car, a = 4 m/s²

The final velocity of the car, v = 28 m/s

The initial velocity of the car, u = 0 m/s (The car starts from rest)

The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;

v = u + a·t

Where;

v = The final velocity of the car, v = 28 m/s

u = The initial velocity of the car = 0 m/s

a = The acceleration of the car = 4 m/s²

t = =The time it will take for the car to reach a velocity of 28 m/s

Therefore, we get;

t = (v - u)/a

t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s

The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.

4 0

2 years ago

The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas

DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length given as

ΔL = L α ΔT

Therefore

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}$

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}$

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0

3 years ago