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3 years ago

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A car traveling 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m

. Part A What was the magnitude of the average acceleration of the driver during the collision?

2 answers:

alexdok [17]3 years ago

8 0

Assuming the acceleration is constant, then we can use the derived equations for rectilinear motion. The equation is written below:

2ax = v²- v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity

Since the car came to a stop, v = 0. Substituting the values,
2a(0.80 m) =0² - [(95 km/h)(1000 m/1 km)(1 h/3600 s)]²
Solving for a,
a = -435.23 m/s²
The sign is negative because it is decelerating.

IceJOKER [234]3 years ago

7 0

The magnitude of the average acceleration of the driver during the collision is about 440 m/s²

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

This problem is about Kinematics.

We will solve it in the following way.

<u>Given:</u>

initial velocity of the car = u = 95 km/h = (475/18) m/s

distance = d = 0.80 m

final velocity of the car = v = 0 m/s

<u>Unknown:</u>

acceleration = a = ?

<u>Solution:</u>

$v^2 = u^2 + 2ad$

$0^2 = (475/18)^2 + 2a(0.80)$

$a = -(475/18)^2 \div 1.6$

$a \approx -440 ~ m/s^2$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Tree

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Answer:

a) For P: $v=0.938\frac{m}{s}$

For Q: $v = 1.876\frac{m}{s}$

b) For P:

$a_{rad}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}=17.60\frac{m}{s^{2}}$

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

$\omega =\frac{\theta}{t}$

One rotation is 360 degrees or 2π radians, so θ=2π

$\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}$

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

$v = \omega R$

for P:

$v = 9.38\frac{rad}{s}*0.1m=0.938\frac{m}{s}$

for Q:

$v = 9.38\frac{rad}{s}*0.2m=1.876\frac{m}{s}$

b) Centripetal acceleration is:

$a_{rad}= \frac{v^2}{R}$

for P:

$a_{rad}= \frac{(0.938)^2}{0.1}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}= \frac{(1.876)^2}{0.2}=17.60\frac{m}{s^{2}}$

c) As seen on a) speed and distance from axis is $v = \omega R$ because ω is constant the if R increases then v increases too.

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3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?

Oksi-84 [34.3K]

The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

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Weight = mg = 0.045 * 10 ≈ 0.45N, g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>

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You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o

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Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

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A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

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3 years ago

A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of

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Answer:

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Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

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1800 m³

mass = 1.29 kg/m³ ( 1800 m³ )

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