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3 years ago

13

You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in

units of m/s^2

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

$0.28 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

$u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s$ is the initial velocity

$v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s$ is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

$a=\frac{2.78-2.22}{2}=0.28 m/s^2$

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Calculate the speed for a car that went a distance of 125 meters in 2 seconds time.

stiv31 [10]

Answer:

<h2>62.5 m/s</h2>

Explanation:

The speed of the car can be found by using the formula

$s = \frac{d}{t} \\$

d is the distance

t is the time

From the question we have

$s = \frac{125}{2} = 62.5 \\$

We have the final answer as

<h3>62.5 m/s</h3>

Hope this helps you

5 0

2 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

3 years ago

When does an object falling vertically through the air reach terminal velocity?

zlopas [31]

Answer:

a. when the acceleration of the objects become negative

7 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

7 0

3 years ago

For a concave mirror, when the object is at focus the image is formed at

Sergeeva-Olga [200]

Answer:

When the object is placed at the focus the image is formed at infinity.

Explanation:

When a ray passes through focus and incident on a concave mirror then it will travel parallel to principal axis after reflection.Hence the image is formed at infinity.

8 0

3 years ago