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Shkiper50 [21]
3 years ago
6

A 2kg toy car traveling at 2m/s strikes a second toy car at rest. The mass of the second toy car is 3. The two stock together an

d move off together. What is their resultant speed?
Physics
1 answer:
Nutka1998 [239]3 years ago
7 0
Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s

Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?

Applying law of conservation of momentum,

Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s

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Which element below has chemical properties similar to the chemical properties of fluorine.
SVETLANKA909090 [29]
For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.
4 0
3 years ago
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
How long will it take to go 150000m traveling at 50km/hr
Tom [10]

Answer:

3000 hurs

Explanation:  just divide 150000 by 50 and get 3000

4 0
3 years ago
Blank is the perceived frequency of a sound wave
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pitch is the perceived frequency of a sound wave.


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