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3 years ago

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A 1500 kg car traveling at 90 km/hr East collides with a 3000 kg truck traveling at 60 km/hr South. They stick together and move

which direction?

1 answer:

EleoNora [17]3 years ago

8 0

Answer:

They both move towards South direction.

Explanation:

NB: The car traveling due east must be backing west and the car traveling due south must be backing north.

From their momentum in calculation below:

Momentum of the car traveling east is MU= 1500x90x1000/3600= 37,500kgm/s.

Momentum of car traveling due South =MU = 3000x60x1000/3600= 50,000. So, from here we see the the momentum of the car moving south has the high momemtum so they stick together and move towards the direction of south.

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That's "<em>DC</em>" . . . Direct Current .

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2 years ago

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The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is

77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

$\% Percent = \frac{V_{nucleus}}{V_{atom}}*100$

$\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100$

$\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100$

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Therefore the percent of the atom's volume is occupied by mass is $7.51*10^{-11}\%$

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2 years ago

A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

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2 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$

$m_1u_1 +m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

$m_2u_2 = (m_1+m_2)v_f$

Rearranging to find the final velocity

$v_f = \frac{m_2u_2}{ (m_1+m_2)}$

$v_f = \frac{ 12500*7.8}{ 12500+7430}$

$v_f = 4.8921ft/s$

The expression for the impulse received by the first car is

$I = m_1 (v-u)$

$I = \frac{W}{g} (v-u)$

Replacing,

$I = \frac{12500}{32.2}(4.89-7.8)$

$I = -1129.65lb\cdot s$

The negative sign show the opposite direction.

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2 years ago

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Mnenie [13.5K]

Answer:

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2 years ago