Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?A. only at positive accelerations

B. only at negative accelerations

C. only at zero acceleration

D. only at constant accelerations

Answers

Answer 1

Answer: If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer is the option D. only at constant accelerations.

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Which air pressure reading is likely an indicator of clear weather?a. 900 mb
b. 950 mb
c. 1000 mb
d. 1050 mb

Answers

1050 mb of air pressure is likely an indicator of clear weather. The correct option among all the options given in the question is option "d". In reality, higher pressure is mostly associated with clear and good weather while stormy or bad weather is often associated with low air pressure.

In a science museum, a 130 kg brass pendulum bob swings at the end of a 14.4 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s. You may want to review (Pages 405 - 407) . Part A At exactly 12:00 noon, how many oscillations will the pendulum have completed

Answers

Answer:

The time in which the pendulum does a complete revolution is called the period of the pendulum.

Remember that the period of a pendulum is written as:

T = 2*pi*√(L/g)

where:

L = length of the pendulum

pi = 3.14

g = 9.8 m/s^2

Here we know that L = 14.4m

Then the period of the pendulum will be:

T = 2*3.14*√(14.4m/9.8m/s^2) = 7.61s

So one complete oscillation takes 7.61 seconds.

We know that the pendulum starts moving at 8:00 am

We want to know 12:00 noon, which is four hours after the pendulum starts moving.

So, we want to know how many complete oscillations happen in a timelapse of 4 hours.

Each oscillation takes 7.61 seconds.

The total number of oscillations will be the quotient between the total time (4 hours) and the period.

First we need to write both of these in the same units, we know that 1 hour = 3600 seconds

then:

4 hours = 4*(3600 seconds) = 14,400 s

The total number of oscillations in that time frame is:

N = 14,400s/7.61s = 1,892.25

Rounding to the next whole number, we have:

N = 1,892

The pendulum does 1,892 oscillations between 8:00 am and 12:00 noon.

Final answer:

The question involves the concept of a simple pendulum whose number of swings is largely influenced by its length and the acceleration due to gravity. By determining the period of the pendulum, one can figure out the number of oscillations over a given time period. The pendulum's damping constant is negligible in determining the number of oscillations.

Explanation:

The subject of this question involves understanding the concept of a simple pendulum and how it relates to harmonic motion. It is widely known that the mass of the pendulum does not influence the oscillations but rather the length of the pendulum wire and acceleration due to gravity are paramount.

First, the necessary step toward calculating the number of swings would be to calculate the period of the pendulum's oscillation. This is given by the formula T=2*π*sqrt(L/g), where L is the length of the pendulum (14.4m) and g is the acceleration due to gravity (~9.81m/s²). Substituting these values will give us the period, T, in seconds.

The pendulum starts swinging at 8:00 am and at 12:00 noon, 4 hours or 14400 seconds will have passed. Therefore the number of oscillations will be calculated by dividing the total time by one period of oscillation.

It is crucial to note that the damping in this instance is quite small and would not significantly affect the number of oscillations.

Learn more about Simple Pendulum here:

brainly.com/question/37947830

#SPJ2

In any one material, all electromagnetic waves have the same A. amplitude.
B. frequency.
C. wavelength.
D. velocity.

Answers

In any one material, all electromagnetic waves have the same speed. But of course, since they may be traveling up, down, left, right, or slanty, they don't all have the same velocity.

I believe it is velocity

The voltage, V (in volts), across a circuit is given by Ohm's law: V=IR, where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance, R, is given by 1R=1R1+1R2. Suppose the current is 3 amps and increasing at 0.02 amps/sec and R1 is 4 ohms and increasing at 0.4 ohms/sec, while R2 is 3 ohms and decreasing at 0.2 ohms/sec. Calculate the rate at which the voltage is changing.

Answers

The rate at which the voltage of the given circuit is changing is gotten to be;

dV/dt = 0.452 V/s

We are given;

Current; I = 3 A

Resistance 1; R1 = 4Ω

Resistance 2; R2 = 3Ω

dR1/dt = 0.4 Ω/s

dR2/dt = 0.2 Ω/s

dI/dt = 0.02 A/s

Now, formula for voltage with resistors in parallel is;

1/V = (1/I)(1/R1 + 1/R2)

Plugging in the relevant values, we can find V;

1/V = (1/3)(1/4 + 1/3)

Simplifying this gives;

1/V = 0.194

Now, we want to find the rate at which the voltage is charging, we need to find dV/dt.

Thus, let us differentiate 1/V = (1/I)(1/R1 + 1/R2) with respect to t to get;

(1/V)²(dV/dt) = [(1/i²)(di/dt)(1/R1 + 1/R2)] + (1/I)[(1/R1²)(dR1/dt) + (1/R2²)(dR2/dt)]

Plugging in the relevant vies gives us;

0.194²(dV/dt) = [(1/3²)(0.02)(¼ + ⅓)] + (⅓)[(1/3²)(0.4) + (1/4²)(0.3)]

>> 0.037636(dV/dt) = 0.001296 + 0.0157

>> dV/dt = 0.016996/0.037636

>> dV/dt = 0.452 V/s

Read more at; brainly.com/question/13539417

Answer:

$(dV)/(dt) = 0.453 Volts/s$

Explanation:

As we know that two resistors are in parallel

so we have

$V = i R$

where we know that

$(1)/(R) = (1)/(R_1) + (1)/(R_2)$

so we have

$(1)/(V) = (1)/(i)((1)/(R_1) + (1)/(R_2))$

now to find the rate of change we have

$(1)/(V^2)(dV)/(dt) = (1)/(i^2)(di)/(dt)((1)/(R_1) + (1)/(R_2)) + (1)/(i)((1)/(R_1^2)((dR_1)/(dt)) + (1)/(R_2^2)((dR_2)/(dt)))$

$(1)/(V) = (1)/(3)((1)/(4) + (1)/(3))$

$(1)/(V) = 0.194$

now from above equation we have

$(0.194)^2(dV)/(dt) = (1)/(3^2)(0.02)((1)/(4) + (1)/(3)) + (1)/(3)((1)/(4^2)(0.4) + (1)/(3^2)(0.2))$

$0.0376(dV)/(dt) = 1.296* 10^(-3) + 0.0157$

$(dV)/(dt) = 0.453 Volts/s$

What force keeps a water spider on the surface of water

Answers

i think it's surface tension force

A thin spherical glass shell in air is filled with an unknown liquid. A horizontal parallel light beam is incident on the shell and it is observed that the light is brought to focus on the surface of the shell directly opposite of the incoming beam. What is the refractive index of the liquid?ie. the answer is 2- I just don't know how they got there