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Alchen [17]
3 years ago
9

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the dist

ance (in ft) traveled before the car comes to a stop? (Round your answer to one decimal place.)
Physics
1 answer:
Bingel [31]3 years ago
4 0

The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance <em>x</em> such that

0² - (73 ft/s)² = 2 (-24 ft/s²) <em>x</em>

==>   <em>x</em> ≈ 111.0 ft

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a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of
Papessa [141]

The initial speed of the bolt is not 58.86 m/s.  

Let a be the acceleration of the rocket.  

During the 4 sec lift off, the rocket has reached a height of  

h = (1/2)*a*t^2  

with t=4,  

h = (1/2)*a^16  

h = 8*a  

Its velocity at 4 sec is  

v = t*a  

v = 4*a  

The initial velocity of the bolt is thus 4*a.  

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,  

h = (1/2)*g*t^2 + V0*t  

Substituting h0=8*a, t=6 and V0=-4*a into it,  

8*a = (1/2)*g*36 - 4*a*6  

Solving for a  

a = 5.52 m/s^2

6 0
3 years ago
How could you prove that
Lubov Fominskaja [6]

Answer:

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Explanation:

5 0
3 years ago
A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p
ella [17]

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

7 0
3 years ago
Wat does gas mean in science
irinina [24]
Basically its just an air like fluid with chemicals.... just like air it just goes as it pleases and fills space
6 0
3 years ago
Read 2 more answers
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
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