Which of the following is a form of technology ?

What do u put on the last blank?

Nookie1986 [14]

The greater the MASS of a moving object, the more kinetic energy it has. <3

8 0

2 years ago

How would you determine how much error there is between a vector addition and the real results

chubhunter [2.5K]

Desired operation: A + B = C; {A,B,C) are vector quantities.

Issue: {A,B} contain error (measurement or otherwise) 

Objective: estimate the error in the vector sum. 

Let A = u + du; where u is the nominal value of A and du is the error in A 
Let B = v + dv; where v is the nominal value of B and dv is the error in B 
Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] 

C = A + B 

w + dw = (u + du) + (v + dv) 

w + dw = (u + v) + (du + dv) 

w = u+v; dw = du + dv 

The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)

6 0

2 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

2 years ago

Which shows increasing entropy?

fomenos

Answer:

mowing a lawn

Explanation:

Entropy is the degree of disorderliness of a system. As a body moves from a more ordered state to a less ordered one, the entropy of the system increases.

Mowing a lawn is a typical example of increasing entropy.
When a lawn is being mowed, the grasses becomes disordered without any fixed orientation.
Folding a clothe is trying to bring orderliness to the clothe patterning.
Washing dishes will make one arrange them in an ordered way.
Falling leaves brings leaves together.

6 0

2 years ago

Read 2 more answers

A 3 kg stone is dropped from a height of 4 m what is its momentum as it strikes the ground

Vitek1552 [10]

Okay first you have to recognize that the maximum Gravitational potential energy will equal the maximum kinetic energy. The maximum GPE will equal when the stone is at its highest point and the max KE will be right before the stone hits the ground. So

GPE max = KE max

mgh = 1/2mv^2

Mass cancels as it is on both sides

gh = 1/2v^2

Multiply by 2

2gh = v^2

Square root

v = √2gh

Now plug in

v = √2(9.8)(4)

v = 8.85 m/s

Now use the mass to calculate the momentum just before it hits the ground as this is the speed right before it hits the ground

p = mv

p = (3)(8.85)

p = 26.55 kgm/s

7 0

2 years ago