Ask question

Ask question

All categories

3 years ago

10

A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

is stationary. The carts stick together after the collision. What is the final kinetic energy of the system?

1 answer:

lina2011 [118]3 years ago

5 0

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

You might be interested in

A proton with mass 1.67*10^-27kg is propelled at an initialspeed of 3.00*10^5m/s directly toward a uranium nucleus 5.00maway. Th

Tresset [83]

Answer:

Explanation:

F = 2.12 x 10⁻²⁶ / x²

Work done by electric field of nucleus

W = ∫ Fdx

= ∫2.12 x 10⁻²⁶ / x² dx

= 2.12 x 10⁻²⁶ ( - 1 / x )

= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )

= - .265 x 10⁻¹⁶ J

1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶

= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶

=( .7515 - .265 )x 10⁻¹⁶

= .4865 x 10⁻¹⁶

.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶

v² = .5826 x 10¹¹

v² = 5.826 x 10¹⁰

v = 2.41 x 10⁵ m /s

b )

Let r be the closest distance

Potential at this point

2.12 x 10⁻²⁶ ( 1 / r )

Kinetic energy

= 0

Total energy = 2.12 x 10⁻²⁶ ( 1 / r )

Total energy at 5 m

= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )

= 7.515 x 10⁻¹⁷ J

So ,

2.12 x 10⁻²⁶ ( 1 / r ) = 7.515 x 10⁻¹⁷

r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷

= .282 x 10⁻⁹

= 2.82 x 10⁻¹⁰ m

c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as

3.00 x 10⁵ m /s

7 0

4 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

An empty capacitor is capable of storing 1.0 × 10-4 J of energy when connected to a certain battery. If the distance between the

dangina [55]

Answer:

Explanation:

Energy stored in a capacitor

= 1/2 C₁V²

capacity of a capacitor

c = εK A / d

k is dielectric and d is distance between plates .

When the distance between the plates is halved and then filled with a dielectric (κ = 4.3)

capacity becomes 4.3 x 2 times

New capacity

C₂ = 8.6 C₁

Energy of modified capacitor

1/2 C₂ V²= 1/2 x 8.6 c x V²

Energy becomes

8.6 times.

Energy stored = 8.6 x 10⁻⁴ J

3 0

4 years ago

Barium chlorate (ba(clo3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this reaction? whe

Margarita [4]

I think the correct answer should be B<span>a(ClO3)2 = BaCl2 + 3O2. In balancing chemical reactions, it important that the number of atoms of each element in each side of the reaction should be the same. Hope this answers the question. Have a nice day.</span>

5 0

4 years ago

Read 2 more answers

Noble gases, such as argon and neon, are known for being extremely non-reactive. Neon and argon are non-reactive because they A.

Greeley [361]

B. have eight electrons in their outer shell

3 0

3 years ago

Read 2 more answers