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Sliva [168]
3 years ago
11

Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.

A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?
Physics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

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Determine the linear velocity of an object with an angular velocity of 5.9 radians per second at a distance of 12 centimeters fr
Andre45 [30]
The linear velocity of a rotating object is the product of the angular velocity and the radius of the circular motion. Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. From the problem, the angular velocity is 5.9 rad per second and the radius is given as 12 centimeters. We calculate as follows:

Linear velocity = angular velocity (radius)
Linear velocity = 5.9 (12 ) = 70.8 cm / s

The linear velocity of the body in motion is 70.8 centimeters per second or 0.708 meters per second.
7 0
3 years ago
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
Who can help me for this question plz:(
Vanyuwa [196]
Force = mass x acceleration

15 = mass x 4

Mass = 15/4

Mass = 3.75 Kg
8 0
3 years ago
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F  = (1200)(0) = 0

4 0
3 years ago
Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc
vivado [14]

Answer:

ΔR_{e} = 84   Ω,     R_{e} = (40 ± 8) 10¹   Ω

Explanation:

The formula for parallel equivalent resistance is

          1 / R_{e} = ∑ 1 / Ri

In our case we use a resistance of each

           R₁ = 500 ± 50  Ω

          R₂ = 2000 ± 5%

This percentage equals

        0.05 = ΔR₂ / R₂

        ΔR₂ = 0.05 R₂

        ΔR₂ = 0.05 2000 = 100   Ω

We write the resistance

        R₂ = 2000 ± 100    Ω

We apply the initial formula

        1 / R_{e} = 1 / R₁ + 1 / R₂

        1 / R_{e} = 1/500 + 1/2000 = 0.0025

        R_{e}  = 400    Ω

Let's look for the error  (uncertainly) of Re

      R_{e} = R₁R₂ / (R₁ + R₂)

       R’= R₁ + R₂

       R_{e} = R₁R₂ / R’

Let's look for the uncertainty of this equation

      ΔR_{e} / R_{e} = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     ΔR_{e} / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     ΔR_{e} / 400 = 0.1 + 0.05 + 0.06

     ΔR_{e} = 0.21 400

     ΔR_{e} = 84   Ω

Let's write the resistance value with the correct significant figures

    R_{e} = (40 ± 8) 10¹   Ω

6 0
3 years ago
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