For an object to slow down and stop due to friction while sliding across the floor which of the following would be true.

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago

How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.

Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0

2 years ago

When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e

Maslowich

When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>. (choice-D)

5 0

2 years ago

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

Svet_ta [14]

Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)

3 0

3 years ago

Use these words in a sentence proton neutron and isotope

Hunter-Best [27]

"The proton and neutron have nothing to do with the isotope little billy"

4 0

3 years ago