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andrey2020 [161]
3 years ago
5

The 2020 Ford police interceptor has a max acceleration of 4.4 m/s2. If a policeperson is sitting on

Physics
1 answer:
mrs_skeptik [129]3 years ago
6 0

Answer:

The policeperson will not be able to catch the speeder in 10 seconds

Explanation:

The maximum acceleration of the 2020 Ford police interceptor, a = 4.4 m/s²

The rate at which the speeding car is moving, v = 70 mph (31.3 m/s)

The time after which the policeperson takes off = 1 second

The time, t, at which the policeperson catches the speeding car is given as follows;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity of the Ford police interceptor = 0

∴ s = 0 × t + 1/2·a·t² = 1/2·a·t²

s = 1/2·a·t²

s = v × (t + 1)

Equating both equations, gives;

v × (t + 1) = 1/2·a·t²

Substituting the values, we get;

31.3 × (t + 1) = 1/2 × 4.4 × t²

31.3·t + 31.3 = 2.2·t²

2.2·t² - 31.3·t - 31.3 = 0

By the quadratic formula, we have;

t = (31.3 ± √((-31.3)² - 4×2.2×(-31.3))/(2 × 2.2)

t ≈ -0.938 seconds or t ≈ 15.165 seconds

The policeperson will catch the speeder in approximately 15.165 seconds

Therefore, the policeperson will not be able to catch the speeder in 10 seconds.

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Answer:

50N

Explanation:

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Which of the following is the energy of motion?
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K.E.=1/2 mv^2

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lapo4ka [179]

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A) psychrometer

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Read 2 more answers
A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.60 m/s. How
kirill115 [55]

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

E_{pot}-E_{friction}=E_{kin}

where:

E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

(56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ]

4 0
3 years ago
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