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3 years ago

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If you're driving 55 MPH and you suddenly need to stop, how many feet will you travel before the car comes to a stop

1 answer:

lys-0071 [83]3 years ago

5 0

ANSWER: 170 Feet
REASON: with good breaks and a dry road your car should stop and skip 170 feet, with perception and reaction time of stopping you should stop within 170 feet

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2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

2 years ago

Help please asap due 20 minutes please help me

Finger [1]

Did you turn it in yet?

4 0

3 years ago

A total distance of 4.5km. The overall journey takes 0.62h, whats the average velocity

Westkost [7]

Answer:

7.3km/hr

Explanation:

v=d/t=4.5km/0.62hrs=7.3km/hr

6 0

3 years ago

You are not harmed by contact with a charged metal ball, even though its voltage may be very high. Is the reason similar to why

Jobisdone [24]

Answer:

.

Explanation:

.

6 0

3 years ago

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer

Mrac [35]

Explanation:

The given data is as follows.

$m_{1}$ = 57 kg, $m_{2}$ = 79 kg

$l_{1}$ = 6.5 m, $l_{2}$ = (6.5 - 1.9) m = 4.6 m

(a) The sum of torque ends about far end is as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5$ = 0

T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b) Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.

8 0

3 years ago