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3 years ago

9

A 1200 kg car traveling at 20.83 m/s comes to a screeching halt in a time of 6.00 seconds. Calculate the force of friction exper

ienced by the car.

1 answer:

Setler [38]3 years ago

4 0

Answer:

4166 N

Explanation:

F x T = 1200 x 20.83

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A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

Liula [17]

The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
   Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D = (1/2) (g) (t²)

Multiply each side by 2 :         2 D =            g    t²

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

   and smaller 'g' ==> longer 't' .--

They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.

Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.

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3 years ago

An electron enters a region of uniform perpendicular en and bn fields. it is observed that thevelocity nv of the electron is una

konstantin123 [22]

<span>v is perpendicular to both E and B and has a magnitude E/B</span>

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3 years ago

Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.

Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

= $\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}$

a = 30 units and b = 70 units

= $\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}$

= 40 units to 100 units

hence, the correct answer is option C which is 50 units.

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3 years ago

A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing

valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

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3 years ago

A person throws a balloon straight down above level ground. The balloon bounces back up into the air. Drag force is significant

love history [14]

Explanation:

a) Balloon is being thrown down and is speeding up;

mg > $F_drag$

b) Balloon is in the air on its way down and moving with constant speed.

$F_drag$ =mg

c) The Balloon is on the ground and rest instantaneously

mg = Normal

d) Balloon is moving slowly downward;

e) Because, at the peak of trajectory drag force is 0.

Drag force = 0

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3 years ago