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tia_tia [17]
3 years ago
11

Two carts with masses of 4.32 kg and 2.66 kg move toward each other on a frictionless track with speeds of 4.86 m/s and 3.93 m/s

respectively. The carts stick together after colliding head-on. Find the final speed.
Physics
1 answer:
ollegr [7]3 years ago
6 0

Answer:

m¹v¹+m¹v¹=(m¹+m²)v

4.32*4.86+2.66*3.93=(3.96+4.86)v

v=31.44/8.82

=3.56

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Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0
3 years ago
B. How can you tell where sugar enters the blood?
Korolek [52]

Answer:

Sugar can’t enter cells directly

Explanation:

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I hope I helped

8 0
3 years ago
Which statement about a right triangle is NOT true?
Alex787 [66]

Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two lengths.

Explanation:

only one that made sense

6 0
3 years ago
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Decreasing the temperature of a reaction decreases the reaction rate because
RoseWind [281]
D) decreasing the temperature lowers the average kinetic energy of the reactants.
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3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

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Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
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