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3 years ago

Please Help!!! Max points!!!!!!

Physics

1 answer:

svp [43]3 years ago

8 0

Answer:

some kind of change

motion

KE = 1/2 m v^2

velocity

drag/air resistance

size

shape

Explanation:

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The motion of a particle along a straight line is described by the equation x=6+4t2 -t 4 , where x is in meter and t is in secon

Aleks [24]

Answer:

The position of the particle is 6m

The velocity of the particle is 16 m/s in negative direction

The acceleration of the object is -40 m/s²

Explanation:

Given;

motion of the particle along a straight line as x = 6 + 4t² - t⁴

The position of the object when t = 2s

x = 6 + 4(2)² - (2)⁴

x = 6 + 16 - 16

x = 6m

The velocity of the object when t = 2s

Velocity = dx/dt

dx/dt = 8t - 4t³

when t = 2s

Velocity = 8(2) - 4(2)³

Velocity = 16 - 32

Velocity = -16m/s

Velocity = 16 m/s (in negative direction)

The acceleration of the object when t = 2s

Acceleration = d²x/dt² = 8 - 12t²

Acceleration = 8 - 12 (2)²

Acceleration = -40 m/s²

5 0

4 years ago

Density of a substance ratio

Inga [223]

Answer:

Density of a substance is the ratio of mass of the substance to its volume.

6 0

2 years ago

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State the law of conversation of momentum

Sliva [168]

The total momentum before a collision is equal to the total momentum after the collision if no external forces act on the system.
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6 0

3 years ago

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$

Put Eq 4 into Eq3

$T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5$

Put Eq 5 into Eq2 and solve for a

$a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6$

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

$T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\$

5 0

4 years ago

Find the kinetic energy of a ball of mass 200 grams moving at a speed of 20 m/s

Aleks04 [339]

Answer:

40 J

Explanation:

$KE = \frac{1}{2} mv^{2} \\KE = \frac{1}{2} (0.200 kg)(20 m/s)^{2} \\KE = 40 J$

6 0

3 years ago

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