Fire trucks have the word FIRE written two ways, as seen here, on the front of the truck. Why is that?

Consider the three displacement vectors

matrenka [14]

Answer:

Explanation:

$\overrightarrow{A} = 3\widehat{i}+3\widehat{j}$

$\overrightarrow{B} = \widehat{i}-4\widehat{j}$

$\overrightarrow{C} = -2\widehat{i}+5\widehat{j}$

(a)

$\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$

$\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}$

$\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}$

Magnitude of $\overrightarrow{D}$ = $\sqrt{2^{2}+4^{2}}$

= 4.47 m

Let θ be the direction of vector D

$tan\theta =\frac{4}{2}$

θ = 63.44°

(b)

$\overrightarrow{E} = - \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$

$\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}$

$\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}$

Magnitude of $\overrightarrow{E}$ = $\sqrt{6^{2}+6^{2}}$

= 8.485 m

Let θ be the direction of vector D

$tan\theta =\frac{6}{-6}$

θ = 135°

4 0

4 years ago

How fast must a 1000 kg car be moving to have a kinetic energy of:

VMariaS [17]

Answer:

$\mathrm{(a)}\: 2\: \mathrm{m/s}\\\mathrm{(b)}\: 20\: \mathrm{m/s}\\$

Explanation:

The kinetic energy of an object is given by $KE=\frac{1}{2}mv^2$ where $m$ is the mass of the object and $v$ is the velocity of the object.

We can set up the following equations with the information given:

$\mathrm{(a)}\: 2.0\cdot 10^3=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$2\: \mathrm{m/s}$}$

For part B, we have the same equation, but kinetic energy is now $2.0\cdot 10^5$ .

Therefore:

$\mathrm{(b)}\: 2.0\cdot 10^5=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$20\: \mathrm{m/s}$}$ .

6 0

3 years ago

The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th

olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0

3 years ago

while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o

sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

$\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}$

The minimum radius not to exceed the centripetal acceleration is 2401 m.

8 0

2 years ago

A particle moves on a line away from its initial position so that after t hours it is s = 6t2 + 2t miles from its initial positi

docker41 [41]

<span>32 mph
First, let's calculate the location of the particle at t=1, and t=4
t=1
s = 6*t^2 + 2*t
s = 6*1^2 + 2*1
s = 6 + 2
s = 8 t = 4
s = 6*t^2 + 2*t s = 6*4^2 + 2*4
s = 6*16 + 8
s = 96 + 8
s = 104
So the particle moved from 8 to 104 over the time period of 1 to 4 hours. And the average velocity is simply the distance moved over the time spent. So:
avg_vel = (104-8)/(4-1) = 96/3 = 32
And since the units were miles and hours, that means that the average speed of the particle over the interval [1,4] was 32 miles/hour, or 32 mph.</span>

6 0

3 years ago