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maria [59]
3 years ago
12

Two cars are driving at the same constant speed on a straight road, with car 1 in front of car 2. car 1 suddenly starts to brake

with constant acceleration and stops in 10 m. at the instant car 1 comes to a stop, car 2 begins to brake with the same acceleration. it comes to a halt just as it reaches the back of car 1. what was the separation between the cars before they starting braking?
Physics
1 answer:
Ganezh [65]3 years ago
4 0
<span> Maths delivers! Braking distance ... If the </span>car<span> is initially travelling at u</span>m<span>/s, then the stopping distance d </span>m<span> ... the </span>speed<span> of the </span>car<span> at the </span>instant<span> the </span>brakes<span> are applied. ... An object with </span>constant acceleration<span> travels the </span>same<span> distance as it would ... We </span>start<span> with the second equation of motion:.</span>
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An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of
blondinia [14]

Answer:

v = 12.12 m/s      

Explanation:

It is given that,

Radius of circle, r = 30 m

The coefficient friction between tires and road is 0.5,

The centripetal force is balanced by the force of friction such that,

v = 12.12 m/s

So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.

5 0
3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
2 years ago
As a result of nuclear fusion, the sun produces what 2 types of energy ??
ELEN [110]

Answer:

heat energy and nuclear energy

Explanation:

as a resukt of nuclear fusion, the sun produces two types of energy.

1. heat energy

2. nuclear energy

3 0
3 years ago
Is this right?? please help me. IT IS SOCIOLOGY!!
Westkost [7]

Answer:

Yes, it is correct : )

Explanation:

Hope this helps!

8 0
2 years ago
Read 2 more answers
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the
OlgaM077 [116]

Answer:

a)   m_v = m_s ((\frac{w_o}{w})² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = \frac{K}{m_s}

         k = w₀² m_s

silicon with virus

         w² = \frac{k}{m_s + m_v}

          k = w² (m_v + m_s)

in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

           w₀²  m_s = w² (m_v + m_s)

           m_v = (\frac{w_o}{w})²  m_s - m_s

           m_v = m_s ((\frac{w_o}{w})² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [(\frac{20.4}{2.85})² - 1)]

          m_v = 1.07 10⁻¹⁴ g

4 0
3 years ago
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