All categories

3 years ago

One of the earliest forms of space management was known as __________ .

1 answer:

3 0

Answer: C
<span>
The Smith System is one of the earliest forms of space management that was invented by Harold Smith. <span> He </span>established the Smith System Driver Improvement Institute to help prevent collisions caused by bad driving habits. The earliest strategies of Smith system includes the following: aim high in steering, keep your eyes moving, get the big picture, make sure others see you and leave yourself an out where you can escape from your current path of travel when potential mistakes on the road happen.</span>

You might be interested in

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

Molodets [167]

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 jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e 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 fina 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 ony 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

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

8 0

3 years ago

an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

4 0

2 years ago

CISTU U

marusya05 [52]

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

To attain the terminal velocity,

Fiscous force=Weight

or, 23.55× eta = 41.6

or, eta = 1.76

whete eta is the coefficient of viscosity.

5 0

3 years ago

The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started

bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

$W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J$

The change in electrostatic potential energy $\Delta U$ , of one point charge q is defined as the product of the charge and the potential difference.

$\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V$

5 0

4 years ago