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valina [46]
3 years ago
9

One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their s

peed by 9.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars
Physics
1 answer:
Leokris [45]3 years ago
3 0

Answer:

v_1 = 7.96 m/s

v_2 = 17.8 m/s

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

K = \frac{1}{2}mv^2

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)

2.5 v_1^2 = 0.5 v^2

v_1 = 0.447 v

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2

2.5(0.447 v + 9)^2 = (v + 9)^2

1.58(0.447v + 9) = v + 9

0.293v = 5.22

v = 17.8 m/s

so speed of first car is

v_1 = 0.447 v = 7.96 m/s

v_2 = 17.8 m/s

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ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
2 years ago
S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum
erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be \frac{4.83598}{c^{\frac{1}{3} } }.

We know that the surface area and volume of the sphere is given by:

A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}

Therefore, the ratio between the surface area and the volume for the sphere will be:

\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}

Equating the volume to the constant c, we will find the value of r.

V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }

Substituting the value of r in the ration between surface area and volume, we get:

\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}

Calculating the constants, we get:

\frac{4.83598}{c^{\frac{1}{3} } }

Hence, the ration between surface area and volume is \frac{4.83598}{c^{\frac{1}{3} } }

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

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3 0
1 year ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
2 years ago
Where would you find the lowest-density seawater?
PolarNik [594]
In rivers that are close to sea
4 0
3 years ago
Cynthia forgot to put the fabric softener in the wash. As her socks tumbled in the dryer, they became charged. If a small piece
polet [3.4K]

Answer:

E = 24000 N/C = 24 KN/C

Explanation:

The electric field experienced by a test charge is given by the following formula:

E = \frac{F}{q}\\\\

where,

E = Electric Field = ?

F = Force of attraction = 3 x 10⁻⁶ N

q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C

Therefore, using these values in the equation, we get:

E = \frac{3\ x\ 10^{-6}\ N}{1.25\ x\ ^{-10}\ C}\\\\

<u>E = 24000 N/C = 24 KN/C</u>

6 0
2 years ago
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