Define: atomic mass unit(amu)

When measuring accelerations with three different graphs (x-t, v-t, a-t) which gives the most reliable results and why?

kipiarov [429]

Answer:

the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

Explanation:

The different graphics depending on time give various information, let's examine what we can get from some

Graph of x -t. from this graph we can obtain the speed through the slope, but the acceleration is not directly obtainable

v-t chart. We can get the acceleration not through the slope and the distance traveled by the area under the curve. Obtaining acceleration is very accurate since it is an average that avoids possible errors in measurements. This is the best graph to find the acceleration

Graph of a-t In this graph the acceleration is a point on the Y axis, it gives some errors because it depends strongly on the possible experimental errors.

In conclusion, the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

5 0

3 years ago

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

$U = \frac{1}{2} kx^2$

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

8 0

3 years ago

A shopping cart is given an initial velocity of 4 m/s and experiences a constant acceleration of 4 m/s2. What is the magnitude o

NNADVOKAT [17]

Answer:

48m

Explanation:

Initial velocity = 4m/s
Acceleration = 4m/s²
Time = 4s
Displacement = ?

From Second equation of motion ,

$\longrightarrow$ s = ut + 1/2at²

$\longrightarrow$ s = 4*4 + 1/2*4*(4)² m

$\longrightarrow$ s = (16 + 2 * 16) m

$\longrightarrow$ s = ( 16 + 32 )m

$\longrightarrow$ s = 48 m

8 0

3 years ago

What is simple machine? write any four advantages if it

Dovator [93]

Answer: See explanation

Explanation:

Simple machine is a machine that has few moving parts or may also have no moving parts and they are usually used to make ones work easier. Examples include wheel and axle, wedge, pulley, lever, screw or an inclined plane.

Their advantages include:

• Less energy is required to perform a particular work.

• It makes work easier.

• It makes work faster.

• Simple machine enable one to do things that might have been difficult to achieve or do.

6 0

3 years ago

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

NeX [460]

Answer:

a) T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ , b) T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$ , d) m₂ = m₁ ( $\frac{ L}{2d} -1$ )

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

∑ τ = 0

T_A d - W₂ x -W₁ L/2 = 0

T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$

b) the tension in string B

we write the expression of the translational equilibrium

∑ F = 0

T_A - W₂ - W₁ - T_B = 0

T_B = T_A -W₂ - W₁

T_ B = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ - g m₂ - g m₁

T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0

m₂ (d-x) = m₁ (0.5 L- d)

m₂ x = m₂ d - m₁ (0.5 L- d)

x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}

$\frac{m_1}{m_2} \ (0.5L -d) = d$