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3 years ago

2. -/1 pointsSerPSE10 2.7.P.017.MI.

1 answer:

8 0

Answer: -7.86cm/s^2 don’t forget negative sign!

v1=14cm/s
dx=2.97-(-5)=7.97cm
t=2.85s

dx= v1*t+1/2at^2

I’m not writing the unit’s into the equation until the end, but you always should include them in all your calculations!! Units are very important and helpful to solve problems! Anyways, here we go;

7.97=(14*2.85)+1/2a(2.85)^2

a=[(7.97)-(14*2.85)]*2/(2.85)^2

a=(-63.86)/(8.1225)

a=-7.86cm/s^2 don’t forget the negative sign!

Any questions please feel free to ask. Thanks.

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Which is produced in a synthesis reaction?

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Answer:

Single Compound

Explanation:

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An example of a synthesis reaction is the reaction that takes place between sodium (Na) and chlorine (Cl) to create table salt.

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2 years ago

A motor requires 400 joules of energy to lift a 5.0 kg mass 2.0 meters. Calculate the efficiency of this motor.

Sauron [17]

Answer:

25%

Explanation:

use F=mg

then use the answer you get from that and plug it into W=Fxh

take that answer and divide it by 400 J and multiply by 100

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2 years ago

A block is attached to a spring, with spring constant k, which is attached to a wall. it is initially moved to the left a distan

Ghella [55]

Answer:

x(t) = d*cos ( wt )

w = √(k/m)

Explanation:

Given:-

- The mass of block = m

- The spring constant = k

- The initial displacement = xi = d

Find:-

- The expression for displacement (x) as function of time (t).

Solution:-

- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.

- We apply the Newton's equation of motion in horizontal direction.

F = ma

-kx = ma

-kx = mx''

mx'' + kx = 0

- Solve the Auxiliary equation for the ODE above:

ms^2 + k = 0

s^2 + (k/m) = 0

s = +/- √(k/m) i = +/- w i

- The complementary solution for complex roots is:

x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]

- The given initial conditions are:

x(0) = d

d = [ A*cos ( 0 ) + B*sin ( 0 ) ]

d = A

x'(0) = 0

x'(t) = -Aw*sin (wt) + Bw*cos(wt)

0 = -Aw*sin (0) + Bw*cos(0)

B = 0

- The required displacement-time relationship for SHM:

x(t) = d*cos ( wt )

w = √(k/m)

3 0

3 years ago

The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

And so the average force exerted on the sprinter is

$F=ma=(66 kg)(2.34 m/s^2)=154.5 N$

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

3 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}

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