How does a bicycles kinetic energy change when it slows down?

Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o

adoni [48]

Answer:

a) v = 0.8 m / s , b) $v_{f}$ = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

Before crash

po = m1 vo1 + m2 vo2

po = m1 vo1

After the crash

$p_{f}$ = (m1 + m2) $v_{f}$

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

K = ½ m v²

v = √2K / m

v = √ (2 32/100)

v = 0.8 m / s

b) The final speed,

p₀ = $p_{f}$

m1 v₀1 = (m1 + m2) $v_{f}$

$v_{f}$ = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

$v_{f}$ = 100 / (3 + 100) 0.8

$v_{f}$ = 0.777 m / s

c) The change in kinetic energy

Initial K₀ = $K_{f}$

K₀ = 32 J

Final $K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.777²

$K_{f}$ = 31.07 J

ΔK = $K_{f}$ - K₀

ΔK = 31.07 - 32

ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

v = √ 2K / m2

v = √ (2 32/3)

v = 4.62 m / s

e) the final speed

v₀₂ = v = 4.62 m/s

p₀ = m2 v₀₂

m2 v₀₂ = (m1 + m2) $v_{f}$

$v_{f}$ = m2 / (m1 + m2) v₀₂

$v_{f}$ = 3 / (100 + 3) 4.62

$v_{f}$ = 0.135 m / s

f) variation of kinetic energy

$K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.135²

$K_{f}$ = 0.9286 J

ΔK = 0.9286-32

ΔK = 31.06 J

4 0

3 years ago

How many basic locomotor skills are there?

CaHeK987 [17]

Eight and I don’t know what else to say but for sure 8

4 0

3 years ago

Fig.4.1

Nutka1998 [239]

Answer:

A

Explanation:

7 0

4 years ago

What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h

Citrus2011 [14]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase of temperature

The sample of silver of our problem has a mass of $m=24.7 kg$ . Its specific heat capacity is $C_s = 236 J/g^{\circ}C$ and the increase in temperature is
$\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C$

Therefore, the amount of heat needed is
$Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J$

8 0

4 years ago

Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine

Flauer [41]

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula: $E=k\frac{q}{r^{2} }$

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= $k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }$

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

q2=q3
they are the same distance from q1

These are actually really nice to have, because now we can simplify our expression to:

$E=k\frac{2q}{r^{2} }$

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.

6 0

3 years ago