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3 years ago

What will the stopping distance be for a 3,000-kg car if -3,000 N of force are applied when the car is traveling 10 m/s?

Physics

1 answer:

Harlamova29_29 [7]3 years ago

8 0

F = force applied to stop the car = - 3000 N

m = mass of the car = 3000 kg

a = acceleration of the car = ?

v₀ = initial velocity of the car before the force is applied to stop it = 10 m/s

v = final velocity of the car when it comes to stop = 0 m/s

d = stopping distance of the car

acceleration of the car is given as

a = F/m

inserting the values

a = - 3000/3000

a = - 1 m/s²

using the kinematics equation

v² = v²₀ + 2 a d

inserting the values

0² = 10² + 2 (-1) d

0 = 100 - 2 d

2 d = 100

d = 100/2

d = 50 m

hence the correct choice is

C. 50 m

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Answer:

The value is $E = 0.06 V$

Explanation:

Generally from the question we are told that

The concentration of $[Zn^{2+}]$ at the cathode is $[Zn^{2+}]_a = 1.6 \ M$

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$E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ]$

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3 years ago

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The third option "I always read the owner’s manual when I purchase a new electrical appliance." It tells us that the customer knows how to properly use the electrical equipment that manipulates and knows the do and do not of such devices, so the risk of electric shock is reduced.

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$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

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$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

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$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

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The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

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