Ask question

Ask question

All categories

3 years ago

9

A cat runs 80 meters to the left in 16 seconds, then 50 meters to the right in 10 seconds, then 70 meters to the left in 14 seco

nds, then 40 m to the right in 8 seconds.
a. What was the distance travelled by the cat?

b. What was the displacement of the cat?

c. What was the average speed of the cat?

d. What was the average velocity of the cat?

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

<h3>Answer:a) Distance= 240m. b) Displacement= 90m. c) Average Speed= 5 $ms^{-1}$ . d) Average Velocity= 1.875 $ms^{-1}$ </h3>

Explanation:

a) Distance ➤ (80+50+70+40) m. ➤ Distance= 240m.

b) Displacement ➤ (50+40)m. ➤ Displacement= 90m

c) Average Speed ➤ Total distance travelled/total time taken

➤ 240/48

➤ 5 $ms^{-1}$

d) Average Velocity ➤ Total displacement/total time taken

➤ 90/48

➤ 1.875 $ms^{-1}$

You might be interested in

Small paragraph explaining how how potential and kinetic energy are related

weeeeeb [17]

Answer:

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment.

Explanation:

4 0

3 years ago

Read 2 more answers

A cat is being chased by a dog. Both are running in a straight line at constant speeds. The cat has a head start of 3.8 m. The

Mars2501 [29]

In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s

7 0

4 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

3 years ago

There is a small scratch on a disco record located 14.5 cm from the center. When played, the scratch makes the record skip 30 ti

LuckyWell [14K]

Answer:

The linear speed of scratch is 45.55 cm/dec

Explanation:

The scratch located from the center at radial distance = 14.5 cm.

Each minute the skip of scratch when disco record played = 30 times

Since, the scratch on the disc start revolving when disc is played and in one minute the scratch revolve 30 times. Therefore, the revolution = 30 RPM

Below is the calculation of linear speed of the scratch.

$Angular \ velocity, \ w \ = \frac{2 \pi \times RPM}{60} \\$

$w = \frac{2 \pi \times 30}{60} = \pi \ rad/dec \\$

$Linear \ velocity = wr \\$

$r = radial \ distance \\$

$v = \pi \times 14.5 \\$

$v = 45.55 cm/dec$

3 0

3 years ago

Describe how the amount of voltage applied to a circuit affects current flow. A) Current and voltage of a circuit are indirectly

Nookie1986 [14]

Increasing the voltage will increase the current proportionately. (C)

That's just Ohm's Law:

<em>"The current through a circuit is directly proportional to the voltage across the circuit. The proportionality constant is the reciprocal of the circuit's resistance."</em>

6 0

3 years ago

Read 2 more answers