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Archy [21]
3 years ago
14

An athlete whose mass is 97.0 kg kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with cons

tant acceleration, a barbell that weighs 420 N N . He lifts the barbell a distance of 0.65 m m in a time of 1.5 s s . Part A Use Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell. Express your answer using two significant figures. F floor Ffloor = nothing N?
Physics
1 answer:
SIZIF [17.4K]3 years ago
6 0

Answer:

470 N.

Explanation:

Using equations of motion:

S = vi*t + 1/2*(a*(t^2))

Given:

S = 0.65 m

t = 1.5 s

vi = 0 m/s

0.65 = 1/2 * (a * (1.5)^2)

a = 1.3/2.25

= 0.578 m/s^2

Force = mass * acceleration due to gravity

= 92 * 0.578

= 53.16 N

Total force = 420 + 53.16

= 473.16 N

= 470 N.

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You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol
kogti [31]

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.  

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

If the solution contains a nonelectrolyte, the current flow is nil and the light bulb does not glow. If the solution contains a strong electrolyte, the current flow is very strong and so the brightness of the light bulb is very high. If the solution contains a weak electrolyte, the current flow is much low compared to the strong electrolyte and the light bulb glows, but the brightness is very low.

3 0
3 years ago
As an interstellar cloud of hydrogen gas shrinks in size, its rate of rotation
FinnZ [79.3K]

Answer:

INCREASES, BECAUSE ITS ANGULAR MOMENTUM IS CONSERVED.

Explanation: Interstellar cloud of Hydrogen is an accumulation of Hydrogen gas in the cloud.

As the Interstellar cloud of Hydrogen shrinks (reduces) in size,the rate of rotation of the shrinked Interstellar cloud Increases because its angular momentum is conserved. GASEOUS MOLECULES MAKE UP ABOUT 99% OF THE INTERSTELLAR CLOUD WITH HYDROGEN HAVING ABOUT 90% OF THE VOLUME OF GASES IN THE INTERSTELLAR CLOUD.

4 0
3 years ago
What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain
Makovka662 [10]

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

d = Path difference

\lambda= wavelength

n = Any integer which represent the number of repetition of the spectrum

In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1

d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

d = 2.5mm

Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm

8 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
A person drives north 3 blocks, then turns east and drives 3 blocks. The driver then turns south and drives 3 blocks. How could
KonstantinChe [14]

Answer:by driving east 3 blocks from the starting point

Explanation:)

4 0
3 years ago
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