(x) A change in position is called:

Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels

True [87]

Answer:

No.

Explanation:

Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?

Considering the formula

V = wr

Where

V = linear speed

W = angular speed

r = radius of the wheel.

But W = 2πrf

Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.

While the radius of the second wheel may be large but it will be of a small frequency.

We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.

7 0

2 years ago

Acceleration problem <br> Show work plz

Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)

vo² = 630 simplifying

vo = 25 m/s result

3 0

3 years ago

Mass (kg) 4.0

densk [106]

Answer:

25 m/s

Explanation:

First of all, we can find the acceleration the object by using Newton's second law of motion:

$F=ma$

where

F = 20.0 N is the net force applied on the object

m = 4.0 kg is the mass of the object

a is its acceleration

Solving for a, we find

$a=\frac{F}{m}=\frac{20}{4}=5.0 m/s^2$

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u = 0 is the initial velocity

$a=5.0 m/s^2$ is the acceleration

t = 5 s is the time

By substituting,

$v=0+(5.0)(5)=25 m/s$

8 0

3 years ago

In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends

Leni [432]

Answer:

angle at which the cutter should be set = 32°

Explanation:

We are given;

Speed of conveyor belt; v1 = 15 cm/s

Width;w = 72 cm

Speed of the cutter that moves across the width; v2 = 24 cm/s

Let's calculate time;

Since the width and the rate at which the cutter works per seconds is given, thus;

Time = w/v2

Time = 72/24 = 3 seconds

Now, we know that, velocity = distance/time

Thus, distance = velocity x time

So, distance covered;d = 15 x 3 = 45 cm

Now, for us to calculate the angle the cutter should be set, we would treat it as a triangle.

When the cutter cuts it, it covers a distance d which is perpendicular to the width.

Thus, to calculate the angle made with the width, we'll use trigonometric ratio.

Thus, d/w = tan θ where θ is the angle at which the cutter should be set.

So, 45/72 = tan θ

tan θ = 0.625

θ = tan^(-1)0.625

θ = 32°

3 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago