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mel-nik [20]
2 years ago
13

What is a safe following distance between your automobile and the vehicle in front of you?

Physics
1 answer:
Colt1911 [192]2 years ago
4 0

Answer:

Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.

Thank you and please rate me as brainliest as it will help me to level up

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What is the science behind the making of pop-up books?
IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0
3 years ago
Plz, Help I just need this question to be answered so I can be done with this worksheet.
Neko [114]

Answer:

2 if I'm not wrong.

I hope it will be useful.

3 0
3 years ago
Mike has a mass of 97 kg. He jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when
Ne4ueva [31]

Answer:

a)  fr = 224.3 N , b)   fr = 224.3 N , c)   v = 198.0  m/s

Explanation:

a) For this exercise let's start by calculating the acceleration in the fall

             v² = v₀² - 2 a (y-y₀)

When it jumps the initial vertical speed is zero

             a = -v² / 2 (y-y₀)

             a = -68 2/2 (1000-2000)

             a = 2,312 m / s²

Let's use the second net law to enter the average friction force

            fr = m a

            fr = 97 2,312

            fr = 224.3 N

b) let's look for acceleration

            v² = v₀² - 2 a y

            a = (v² –v₀²) / 2 (y-y₀)

            a = (4² - 68²) / 2 (0-1000)

            a = 2,304 m / s²

            fr = m a

            fr = 97 2,304

            fr = 223.5 N

c) the speed of the wallet is searched with kinematics

           v² = v₀² - 2 g (y-y₀)

           v = √ (0-2 9.8 (0-2000))

           v = 198.0  m/s

4 0
3 years ago
Please help <br>problems 2a.,2b.,3a.,and 3b.​
Darina [25.2K]

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}

8 0
3 years ago
a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe
Mariana [72]

Answer:

1.34352 kg

Explanation:

m_w = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

\Delta T = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

m_v = Mass of water in the vessel

Here the potential energy will balance the internal energy

m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg

Mass of the water in the vessel is 1.34352 kg

6 0
3 years ago
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