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3 years ago

What is the displacement of an object that moves from the origin to a position at −12 m?

Physics

2 answers:

bixtya [17]3 years ago

8 0

Displacement is the shortest distance travelled...

hence in this case the displacement will simply be 12m

Mrrafil [7]3 years ago

6 0

Answer:

Displacement, d = 12 meters

Explanation:

The displacement of an object that moves from the origin to a position at −12 m is 12 meters.

The distance from the origin to the maximum position is called displacement of an object. The maximum displacement of the particle in a wave is called crest and the minimum displacement is called the trough. -12 shows the displacement in negative axis.

So, the displacement of an object is 12 meters.

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Goshia [24]

Answer:

The output out be 200

Explanation:

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3 years ago

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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

3 years ago

Which statements did both Aristotle and Ptolemy assume? Check all that apply.

tensa zangetsu [6.8K]

Bodies in space traveled in circles.

The planets revolved around the Earth.

6 0

3 years ago

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The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the

ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0

3 years ago

if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,

ipn [44]

False.

The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
$U=mgh=(0.2 kg)(9.81 m/s^2)(0.3 m)=0.6 J$

This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.

7 0

3 years ago