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Triss [41]
3 years ago
7

A sled of mass 8 kg slides along the ice. It has an initial speed of 4 m/s but

Physics
1 answer:
marysya [2.9K]3 years ago
6 0

As our story begins, the sled ... whose mass is 8 kg ...  is sliding along the ice at a speed of 4 m/s.

The sled's kinetic energy is (1/2 m v²) = (4 kg · 16 m²/s²) = 64 J .

After what seems like only the blink of an eye, the sled is no longer sliding.  It is stationary.  Motionless.  At Rest.  Just sitting there !  

Its speed has been reduced to zero and ... because kinetic energy is the energy of motion ... the sled's kinetic energy is now also zero.  Sixty-four Joules of energy have disappeared !

How can this be ? ! ? We know that energy is conserved.  It can never just appear out of nothing, and it can never just disappear into nothing.  If energy suddenly appears, it had to come from somewhere, and if energy suddenly disappears, it had to go somewhere.  So where did our 64 Joules of kinetic energy go ?

It went into the ice, THAT's where !  We can say that the sled did 64J of work, and melted a thin slick layer of water on the surface of the ice.  OR we can say that friction did NEGATIVE 64J of work on the sled, to cancel the 64J that it had originally, sap its kinetic energy, and bring it to rest.

I think <em>choice-B</em> was supposed to say "<em>B. -64J</em>", but somebody typed it sloppily and neglected to proofread it before posting.

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What is the concentration of OH– ions at a pH = 6?
makkiz [27]
<h3><u>Answer;</u></h3>

1 × 10^-8 M

<h3><u>Explanation</u>;</h3>

pH is given by the -log[H+] while

pOH is given by the -log[OH-]

But;

pH + pOH = 14

Thus; if pH is 6, then pOH = 8

pOH = 8

-log[OH-] = 8

[OH-] = 10^-8 M

The concentration of OH- ions at a pH of 6 is 1 × 10^-8 M

8 0
3 years ago
If you kicked your mom <br><br> would she be mad?
vfiekz [6]

Answer:

Yes !

Explanation:

8 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
One of the longer sides of a kite is 39 feet. one of the shorter sides is 25 feet. half the length of the shorter diagonal is 15
pychu [463]
The area of a triangle is found by multiplying the height of the triangle by the length of the base and dividing them both by 2. The length of the shorter side in the equation is useless information, so just multiply 39 by 25 and divide that by 2. A=487.5 sq ft. Also, that's a pretty big kite. 
8 0
3 years ago
A 210 kg meteoroid is heading toward the earth accelerates by 2.4 × 105 m/s2.
lidiya [134]
According to Newton's second law of motion, Force is the product of mass and acceleration of the object.
So, F = m * a

Here, m = 210 Kg
a = 2.4 * 10⁵ m/s²

Substitute their values, 
F = 210 * 2.4 * 10⁵ N
F = 504 * 10⁵ N
F = 5.04 * 10⁷ N

In short, Your Answer would be Option B

Hope this helps!
6 0
3 years ago
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