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Flura [38]
2 years ago
6

If the velocity of an object is zero at one instant, what is true about the acceleration of that object

Physics
1 answer:
Charra [1.4K]2 years ago
8 0

Answer:

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

Explanation:

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

If the velocity of an object is zero at one instant, then the acceleration of that object may have the following characteristics;

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

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What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two
lesya [120]

Answer:

Power  = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

\beta (dB)= 10 \  log_{10 } (\dfrac{I}{I_o})

where;

the threshold of hearing   I_o = 10^{-12} (W/m^2)

\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})

10^{10.2} =  \dfrac{I}{10^{-12}}

I = 10^{10.2} \times 10^{-12}

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power  = 124.50 W

7 0
3 years ago
A pulling force is called what?<br> A. normal<br> B. tension<br> C. balanced<br> D. compression
joja [24]
<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
3 0
3 years ago
Read 2 more answers
An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh
Katen [24]

If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.

The  graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.

  • A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m

For more information on work done, visit

brainly.com/subject/physics

5 0
2 years ago
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An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0
2 years ago
NEED HELP ASAP
postnew [5]

Explanation:

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