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2 years ago

The first step of the scientific method is

Physics

2 answers:

Law Incorporation [45]2 years ago

6 0

Answer: Objective Observations

Explanation: The first step in the Scientific Method is to make objective observations. These observations are based on specific events that have already happened and can be verified by others as true or false.

Ray Of Light [21]2 years ago

3 0

Answer: make objective observations.

Explanation:

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A sand core used to form the internal surfaces of a steel casting experiences a buoyancy force of 225 n. the volume of the mold

pashok25 [27]

Newton’s second law gives us the relationship of force F, mass m and acceleration a. The formula is given as:

F = m a --> 1

However we also know that the relationship of mass m, density ρ, and volume V is:

m = V ρ --> 2

Therefore substituting equation 2 to equation 1:

F = ρ V a = ρ V g

where a is acceleration due to gravity, ρ is density of water and V is the volume of the casting, therefore:

F = (1x10^-3 kg/cm^3) (4840 cm^3) (9.8 m/s^2)

F = 47.432 kg m/s^2

F = 47.432 N

Going back to equation 1:

47.432 N = m (9.8 m/s^2)

m = 4.84 kg

Hence the weight of the final casting is 4.84 kg

5 0

3 years ago

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th

Diano4ka-milaya [45]

Answer:

$\frac{K.E_r}{K.E}=2.875$

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

$K.E_r=(\gamma-1)mc^2$ ...........(1)

where,

$\gamma$ = relativistic factor, given as; $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Now, the classical kinetic energy is given as:

$K.E = \frac{1}{2}mv^2$ ..........(2)

Dividing the equation (1) by (2) we get

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}$

$\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}$

substituting the values in the equation we get,

$\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}$

$\frac{K.E_r}{K.E}=2.875$

5 0

3 years ago

A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reac

Step2247 [10]

Answer:

$h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}$

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t Equation (1)

Where:

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀ : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y² $h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}$

2gh= (v₀*sin θ₀)²

$h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}$

7 0

2 years ago

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

goldfiish [28.3K]

Answer:

2.55348650884 m/s

2.67400481907 m/s²

10.6960192763 m/s²

Explanation:

d = Diameter of the ride = 16 ft= $16\times 0.3048=4.8768\ m$

r = Radius = $\dfrac{4.8768}{2}=2.4384\ m$

t = Time taken = 6 seconds

Velocity is given by

$v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 2.4384}{6}\\\Rightarrow v=2.55348650884\ m/s$

The speed is 2.55348650884 m/s

Acceleration is given by

$a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{6})^2}{2.4384}\\\Rightarrow a=2.67400481907\ m/s^2$

The acceleration is 2.67400481907 m/s²

When the speed is halved

$v=\dfrac{2\pi 2.4384}{\dfrac{6}{2}}=5.10697301768\ m/s$

$a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{\dfrac{6}{2}})^2}{2.4384}\\\Rightarrow a=10.6960192763\ m/s^2$

The acceleration is 10.6960192763 m/s²

3 0

3 years ago

A proton is trapped in a one-dimensional well that is 200 pm wide. What is the ground state energy of the proton, if the potenti

Harrizon [31]

Answer:

E=1.50\times 10^{-18}J

Explanation:

Energy of the one dimensional infinite well,

$E=\frac{n^{2}h^{2} }{8mL^{2} }$

Given that, $L=200 pm\\L=200\times 10^{-12}m$

For the ground state n=1,

Therefore energy is,

$E=\frac{(6.626\times 10^{-34})^{2}}{8(9.1\times 10^{-31})(200\times 10^{-12}) ^{2} }\\E=1.50\times 10^{-18}J$

8 0

3 years ago