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Alla [95]
3 years ago
6

What is scientific inquiry (method) ?

Physics
1 answer:
Shtirlitz [24]3 years ago
6 0
A method of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses
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Difference between hair dryer and heat gun
Bumek [7]
The heat gun<span> obviously wins this round. Master Appliance </span>heat guns<span> can reach temperatures of up to 1,000 Fahrenheit. A handheld </span>blow dryer<span> might reach 131 degrees Fahrenheit. A </span>hair dryer<span> gets hot enough to burn skin, but not hot enough to complete serious tasks like striping paint and removing serious. By the way I got this from google.</span>
5 0
3 years ago
Sam lifted his backpack with 5 Newtons of force a total of 400
irga5000 [103]

2000joules

Explanation:

work done=force×meters

4 0
3 years ago
The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne
Lisa [10]

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

f=N*miu

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

F - f = m*a

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]

6 0
3 years ago
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

6 0
2 years ago
A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the
laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, \Delta f = 4\ beats/s

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

\Delta f = f - f_{i}

f_{i} = f \pm \Delta f

when

f_{i} = f + \Delta f = 523 + 4 = 527 Hz

when

f_{i} = f - \Delta f = 523 - 4 = 519 Hz

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0
3 years ago
Read 2 more answers
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