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Nimfa-mama [501]
3 years ago
12

What is the kinetic energy of a 1700 kg car traveling at a speed of 30 m/s (â65 mph)? does your answer to part b depend on the c

ar's mass?
Physics
1 answer:
MArishka [77]3 years ago
6 0
The kinetic energy of an object is given by
K= \frac{1}{2} mv^2
where m is the mass of the object and v its velocity.

For the car in the problem, m=1700 kg and v=30 m/s, so the kinetic energy of the car is
K= \frac{1}{2}mv^2= \frac{1}{2}(1700 kg)(30 m/s)^2=7.65\cdot 10^5 J

and as we can see, yes, the answer depends on the car's mass.
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Answer:

90 kJ

Explanation:

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W = (900 N) (100 m)

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W = 90 kJ

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Answer:

3 ohms

Explanation:

6×6/6+6 =3 .............

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3 years ago
A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th
Soloha48 [4]

Answer:

The velocity of mass 2m is  v_B = 0.67 m/s

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = v_1 =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = v_A

Let The finial speed  of the billiard ball  B = v_B

      According to the law of conservation of Energy

                 \frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

              Substituting values  

                \frac{1}{2} m (1)^2  = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

Multiplying through by \frac{1}{2}m

                1 =v_A^2 + 2 v_B ^2 ---(1)

    According to the law of conservation of Momentum

            mv_1 + 2m(0) = mv_A + 2m v_B

    Substituting values

            m(1)  = mv_A + 2mv_B

Multiplying through by m

           1 = v_A + 2v_B ---(2)

making v_A subject of the equation 2

            v_A = 1 - 2v_B

Substituting this into equation 1

         (1 -2v_B)^2 + 2v_B^2 = 1

         1 - 4v_B + 4v_B^2 + 2v_B^2 =1

          6v_B^2  -4v_B +1 =1

          6v_B^2 -4v_B =0

Multiplying through by \frac{1}{v_B}

          6v_B -4 = 0

            v_B = \frac{4}{6}

            v_B = 0.67 m/s

4 0
3 years ago
When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm .
vlada-n [284]

Answer:

m = 56.5 kg

Explanation:

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Δm*g / Area = p * g * Δh   ..... Eq1

m = \frac{p*h*A}{g}

m = \frac{850 * 0.4 * pi*0.46^2}{4*9.81}\\\\m= 56.5 kg

8 0
2 years ago
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