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hjlf
3 years ago
14

What happens to the acceleration of an object if the net force on it increases

Physics
2 answers:
mart [117]3 years ago
7 0

Answer:

The acceleration increases is the answer

Explanation:

Virty [35]3 years ago
6 0
IS any one going to answer this I need the answer also
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A teacher told a learner to react benzene (CH) with chlorine (Cl₂) to
kolezko [41]

The minimum quantity of benzene, C₆H₆ needed for the reaction is 106.67 g

<h3>How to determine the theoretical yield of chlorobenzene, C₆H₅Cl</h3>

From the question given, the following data were obtained

  • Actual yield = 100 g
  • Percentage yield = 65%
  • Theoretical yield =?

Percentage yield = (Actual / Theoretical) × 100

65% = 100 / Theoretical

0.65 = Actual / Theoretical

Cross multiply

0.65 × Theoretical = 100

Divide both sides by 0.65

Theoretical = 100 / 0.65

Theoretical yield = 153.85 g

<h3>How to determine the mass of benzene, C₆H₆ needed</h3>

Balanced equation

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Molar mass of C₆H₆ = 78 g/mol

Mass of C₆H₆ from the balanced equation = 1 × 78 = 78 g

Molar mass of C₆H₅Cl = 112.5 g

Mass of C₆H₅Cl from the balanced equation = 1 × 112.5 = 112.5 g

SUMMARY

From the balanced equation above,

112.5 g of C₆H₅Cl were obtained from 78 g of C₆H₆

Therefore,

153.85 g of C₆H₅Cl will be produced from = (153.85 × 78) / 112.5 = 106.67 g of C₆H₆

Thus, the minimum amount of benzene, C₆H₆ needed for the reaction is 106.67 g

Learn more about stoichiometry:

brainly.com/question/16735180

#SPJ1

3 0
2 years ago
How are theories and laws connected ​
Anni [7]

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

4 0
2 years ago
Read 2 more answers
A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

8 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
2 years ago
Which of the following is NOT part of a circuit?
uysha [10]
I believe it’s a. Rim
5 0
2 years ago
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