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2 years ago

Mixing baking soda with vinegar is an example of an oxidation-reduction

Physics

1 answer:

34kurt2 years ago

4 0

Answer:

B. False

Explanation:

The reaction of baking soda and vinegar produces carbon dioxide gas. It is an example of precipitation reactions.

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Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.2min to turn thr

bearhunter [10]

Sure.
Can I use your answer to part-'a' ?

If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of

   (32 rev/min²) x (1.2 min) = 38.4 rev/min .

Check:

If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is

   (1/2) (0 + 38.4) = 19.2 rpm average

At an average speed of 19.2 rpm for 1.2 min,
it covers

   (19.2 rev/min) x (1.2 min) = 23.04 revs .

That's pretty close to the "23" in the question, so I think that
everything here is in order.

4 0

2 years ago

Read 2 more answers

Which two particles are present in the nucleus of an atom? Electrons and neutrons Electrons and molecules 1 TH O Protons and neu

aleksandrvk [35]

Answer:

Protons and neutrons

Explanation:

8 0

2 years ago

An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

For m = 7 [g] and V = 2.3 [cm³]

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

For m = 10 [g] and V = 7 [cm³]

$Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$

For m = 15 [g] and V = 5 [cm³]

$Ro=15/5\\Ro=3[g/cm^{3} ]\\$

For m = 21 [g] and V = 8 [cm³]

$Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$

5 0

2 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

A person takes a trip, driving with a constant speed of 94.5 km/h except for a 22.0 min rest stop. If the person's average speed

qwelly [4]

From the average speed you can fix an equation:

Average speed = distance / time

You know the average speed = 65.1 kg / h, then

65.1 = distance / total time,

where total time is the time traveling plus 22.0 minutes

Call t the time treavelling and pass 22 minutes to hours:

65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1

From the constant speed, you can fix a second equation

Constant speed = distance / time traveling

94.5 = distance / t ==> distance = 94.5 * t

The distance is the same in both equations, then you have:

[t +22/60] * 65.1 = 94.5 t

Now you can solve for t.

65.1t + 22*65.1/60 = 94.5t

94.5t - 65.1t = 22*65.1/60

29.4t = 23.87

t = 23.87 / 29.4

t = 0.812 hours

distance = 94.5 km/h * 0.812 h = 76.7 km

Answers: 1) 0.81 hours, 2) 76.7 km

4 0

3 years ago