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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

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

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

3 years ago

Read 2 more answers

A point moves on the x-axis in such a way that its velocity at time t (t > 0) is given by v=ln t/t . At what value of t does

Olenka [21]

Answer:

Explanation:

Given

Velocity of point is given by $v=\frac{\ln t}{t}$

To get maximum or minimum velocity differentiate v w.r.t t

$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\frac{1}{t}\times t-1\times \ln(t)}{t^2}$

so $1-\ln (t)$ should be equal to zero

$\ln (t)=1$

$t=e$

i.e. $t=2.718\ s$

5 0

3 years ago

A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)

loris [4]

Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s

7 0

3 years ago

Physics is really confusing...is there anyone that can show it to me?

coldgirl [10]

Answer:

1.0×10³ N

Explanation:

μs is the static coefficient of friction. That's the friction that acts on a stationary (non-moving) object when being pushed or pulled.

μk is the kinetic coefficient of friction. That's the friction that acts on a moving object.

To budge the pig (while it's still stationary), we need to overcome the static friction.

F = N μs

For a non-moving object on level ground, the normal force N equals the weight.

F = mg μs

Given m = 130 kg and μs = 0.80:

F = (130 kg) (9.8 m/s²) (0.80)

F = 1019.2 N

Rounded to two significant figures, the force needed to budge the pig is 1.0×10³ N.

3 0

3 years ago

Objects A and B are brought close to each other. Object A will soon become positively charged. Identify the charge that must tra

Rzqust [24]

B because that’s why they are going out of the circle

4 0

3 years ago