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Ad libitum [116K]
3 years ago
5

A driver of a car traveling 12 m/s applies the breaks. After 2 seconds the speed drops to 8 m/s what is the cars acceleration?

Physics
2 answers:
Afina-wow [57]3 years ago
6 0

Answer:

Deceleration of the car is 2\frac{m}{s^{2} }.

Explanation:

Given:

initial speed = 12 m/s

final speed = 8 m/s

time = 2 s

To find:

acceleration = ?

Formula used:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

Solution:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

8 = 12 + 2 a

-4 = 2 a

a = -2 \frac{m}{s^{2} }

The negative sign implies that it is deceleration. Thus, deceleration of the car is 2\frac{m}{s^{2} }.



Minchanka [31]3 years ago
5 0

Answer:

Acceleration value of car = -2 m/s^2

Explanation:

 We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Here a car is traveling at 12 m/s, so initial velocity = 12 m/s

After 2 seconds the speed drops to 8 m/s, so final velocity = 8 m/s

Time taken = 2 seconds

Substituting

        8 = 12 + a * 2

         2a = -4

         a = -2 m/s^2

 Acceleration value of car = -2 m/s^2

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3 years ago
A 1.5 kg frictionless pendulum is released from point A at an angle θ of 35 degrees. The speed of the pendulum at point C is 3.4
sergeinik [125]

The length of the pendulum is 3.3 m.

The given parameters:

  • Mass, m = 1.5 kg
  • Angle, θ = 35⁰
  • Speed, v = 3.4 m/s

<h3>What is principle of conservation of energy?</h3>
  • The principle of conservation of energy states that, the total energy of a system is always conserved.

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

g(L - Lcosθ) = ¹/₂v²

gL(1 - cosθ) = ¹/₂v²

L = \frac{v^2}{2g(1- cos\theta)} \\\\L = \frac{(3.4)^2}{2\times 9.8(1 - cos35)} \\\\L = 3.28 \ m\\\\L = 3.3 \ m

Thus, the length of the pendulum is 3.3 m.

Learn more about length of pendulum here: brainly.com/question/8168512

3 0
3 years ago
The graph is tied to the reading
AleksandrR [38]
I. Positive acceleration increases velocity. Negative acceleration decreases velocity. runner A sped up until the finish line and then slowed to a stop.

ii. Zero a acceleration implies a constant, unchanging velocity not a zero velocity. runner B achieved some velocity prior to 8s and is moving and must slow down to reach a stop.

iii. None. No aspects of this reasoning are correct. Everything she says is wrong. See iv for what/why.

iv. The sign on acceleration denotes the direction of *change in velocity* not change in direction. The sign on velocity can denote change in direction but only “forward” or “reverse” along a particular path. Cardinal direction is not indicated, generally, by the sign on velocity. It may correspond to North/South situationally but it is not an built-in feature of velocity and its sign. For example, if you are traveling with positive velocity and turn left to continue your journey you still have a positive velocity in the new direction. In fact, if you turn left again, traveling in the opposite direction as the one you started with your velocity would still be positive… in the new direction. The velocity relative to original direction could be said to be negative but that would be a confusing way to describe a journey. Maybe if you stopped the vehicle and moved in reverse, you could meaningfully say velocity was negative.
5 0
3 years ago
A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle
Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

\tau=NIABsin\theta   (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

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A machine is designed to lift an object with a weight of 12 newtons. if the input force for the machine is setat 4 newtons, what
Alja [10]
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