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3 years ago

7

A 2000 kg car experiences a constant braking

1 answer:

bezimeni [28]3 years ago

5 0

Solving for acceleration:

F = ma
10000 = 2000*a
a = 5 m/s^2

Solving for velocity:

Vf = Vi + a(t)
0 = Vi - (5)(6)
Vi = 30 m/s

Solving for displacement or distance:

S = Vi*t - 1/2 * a * t^2
S = 30(6) - 1/2 (5)(6)^2
S = 90m

The car would have traveled 90m before coming to rest.

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The molecules in which of the following have the greatest average molecular kinetic energy?

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We would need examples but kinetic energy is the energy you get when something moves. The answer will be the one where the molecules are more spread out because they move more.

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4 years ago

Stacy's home is sprayed with a pesticide to control spiders. She didn't see any spiders for a period of time, but eventually, sh

kirza4 [7]

A) The spiders with pesticide-resistant traits survived & reproduced.

The spiders that survived the pesticide passed on their genes to their offspring, and when their offspring came back to Stacy's house, they had the resistance to pesticide because of their parents.

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3 years ago

An auditorium measures 35.0 m x 30.0 m x 5.0 m. The density of air is 1.20 kg/m^3. (a) What is the volume of the room in cubic f

Ksenya-84 [330]

Answer:

(a) 1852259 $ft^3$ (b) 489085.47 pound

Explanation:

We have given auditorium measures 35 m×30 m×5 m

We know that 1 meter = 3.28 feet

So the measure of auditorium = 35×3.28 feet ×30×3.28 feet× 5×3.28 feet

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Density is given as $d=1.20kg/m^3$

(b) weight of air = volume × density $=185259.648\times 1.2=222311.577kg$

We know that 1 kg = 2.20 pound

So 222311.577 kg =222311.577×2.20=489085.47 pound

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4 years ago

a 1.2 kg rocket is launched from ground level with initial velocity of 12 m/s. What is the maximum height the rocket can reach?

Tanzania [10]

As the rocket is launched from the ground its height will go on increasing till it stops

So the height of rocket will be maximum when its speed becomes zero

so here we can use energy conservation theory

$PE = KE$

$mgH = \frac{1}{2}mv^2$

$9.8*H = \frac{1}{2}*12^2$

$H = 7.35 m$

So it will reach upto height 7.35 m

5 0

3 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

4 years ago