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3 years ago

12. In which of the positions shown in the picture, will Joshua have both potential and kinetic energy?

Physics

1 answer:

Zielflug [23.3K]3 years ago

8 0

Answer:

Answer: D. Only R

Explanation:

The Principle Of Conservation Of Mechanical Energy

In the absence of friction, the total mechanical energy is conserved. That means that

Em=U+K is constant, being U the potential energy and K the kinetic energy.

At the top of the path (point Q), Joshua is at rest but his height is at a maximum. Thus his kinetic energy is zero.

At point P, Joshua is at the minimum level (potential energy zero) and his speed is at a maximum.

At point R, Joshua has some height and some speed which means he has both types of energy.

Answer: D. Only R

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The athlet at point A runs 150m east, then 70m west and then 100 east

denpristay [2]

Answer:

180m to the east

Explanation:

Displacement is the distance traveled in a specific direction. It is a vector quantity with both magnitude and direction. Therefore, the start and finish position is very paramount.

point A runs 150m east,

70m west

100m east

150m

--------------------------------------------------------→

70m

←---------------------

100m

-----------------------------------→

The displacement of the athlete = 150 - 80 + 100 = 180m to the east

6 0

3 years ago

Please help.

Alex787 [66]

Explanation:

mathematically,

$kinetic \: energy \: = \frac{1}{2} m {v}^{2}$

where m= mass

and v=velocity

given that,

kinetic energy=493900J

mass =5040kg

making velocity the subject

$v = \sqrt{ \frac{2 \times kinetic \: energy}{mass} }$

substitute their values in the formula

$v = \sqrt{ \frac{2 \times 493900}{5040} }$

$v = \sqrt{ \frac{987800}{5040} }$

$v = \sqrt{196}$

$v = 14m {s}^{ - 1}$

Thus the velocity is 14meters per second

6 0

3 years ago

A projectile is launched horizontally from a cliff top at 12 m/s. Determine the x-y positions at 1-second intervals. The launch

Natalka [10]

The x position will be "12 m" and the y position will be "-4.9 m".

According to the question,

Along x direction, Initial velocity:

12 m/s

Along x direction, Acceleration is:

Along y direction, Acceleration is:

-10 m/s²

Time,

t = 1 second

→ The X coordinate will be:

= $12\times t$

= $12\times 1$

= $12 \ m$

→ The Y coordinate will be:

= $-(u_y t+\frac{1}{2}a_y t^2 )$

= $0+4.9 t^2$

= $4.9 \ t^2$

= $-4.9\times 1^2$

= $-4.9 \ m$

Thus the above answer is correct.

Learn more:

brainly.com/question/17108357

5 0

3 years ago

A torque of 4 Nm is required to rotate the intermediate cylinder at 30 radians/min. Calculate the viscosity of the oil. All cyli

NISA [10]

Answer: viscosity of the oil becomes μ = 0.2003 pa.s

Explanation:

from the given question, we have that

torque T = 4Nm

height H = 450mm = 0.45m

rotational speed ω = 30rpm = 30×2π/60 = 3.14 rad/sec

also to calculate for the linear velocity of the intermediate cylinder,

V = Rω

where R is the radius of the cylinder.

now we substitute values of R and ω, we have

V = Rω

V = 0.15 × 3.14 = 0.471 m/s

calculating the drag force on the two side of the cylinder;

Fd = 2μA (Δv/Δy)

Fd = 2μA V/h

where A = 2πRH

Fd = 2μ(2πRH) V/h

h = thickness, μ = viscosity of oil, H = height of cylinder, and V = mean velocity of cylinder

substituting values we have,

Fd = 2μ * (2π*0.15*0.45)* 0.471/0.003

Fd = 133.10μN

from the torque equation, we can calculate the viscosity

thus, T = Fd*R

we already have Fd = 133.10μN, substituting it into the above expression we have

T = 133.10μ * 0.15 where T is 4Nm,

∴ μ = 4/ 133.1*0.15

μ = 0.2003 pa.s

i hope this helps, cheers

6 0

3 years ago

If a net force is greater than 0 Newtons, what is the force?

goldenfox [79]

Answer:disequilibrium

Explanation:

When the net force is not zero it is called disequilibrium

5 0

4 years ago

12. In which of the positions shown in the picture, will Joshua have both potential and kinetic energy?​

12. In which of the positions shown in the picture, will Joshua have both potential and kinetic energy?