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2 years ago

7

Tyler has learned that potential energy is energy stored. Tyler's teacher asks the students to come up with a demonstration of p

otential energy. Which of the following would best work for this demonstration?
A

A parked car in the driveway

B

Wheels spinning on the ground

C

A rock after it has been thrown

D

A child at the top of a slide

2 answers:

Advocard [28]2 years ago

7 0

D.A child at the top of a slide

MrMuchimi2 years ago

3 0

D; cause potential energy is when something has the potential to move

You might be interested in

Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assu

Vera_Pavlovna [14]

Answer:

The partial pressure of H2 is 0.375 atm

The partial pressure of Ne is also 0.375 atm

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

Pressure of sealed container = 1.5 atm

Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

7 0

2 years ago

A 94-ft3/s water jet is moving in the positive x-direction at 18 ft/s. The stream hits a stationary splitter, such that half of

vitfil [10]

Answer:

FR<em>x </em>= 960.37 lbf (←)

FR<em>z </em>= 0 lbf

Explanation:

Given:

Q = 94 ft³/s

vx = 18 ft/s

ρ = 62.4 lbm/ft³

∅ = 45°

<em>Assumptions: </em>

1. The flow is steady and incompressible.

2 . The water jet is exposed to the atmosphere, and thus the pressure of the water jet before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces.

3. The gravitational effects are disregarded.

4. The flow is nearly uniform at all cross sections, and thus the effect of the momentum-flux correction factor is negligible, β ≅ 1.

<em>Properties:</em> We take the density of water to be ρ = 62.4 lbm/ft³

Analysis: The mass flow rate of water jet is

M = ρ*Q = (62.4 lbm/ft³ )(94 ft³/s) = 5865.6 lbm/s

We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate <em>M</em>). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z.

The momentum equation for steady flow is

∑ F = ∑ (β*M*v) <em>out</em> - ∑ (β*M*v) <em>in</em>

We let the x- and y- components of the anchoring force of the splitter be FR<em>x</em> and FR<em>z, </em>and assume them to be in the positive directions. Noting that

v₂ = v₁ = v and M₂ = (1/2) M, the momentum equations along the x and z axes become

FR<em>x </em>= 2*(1/2) M*v₂*Cos ∅ - M*v₁ = M*v*(Cos ∅ - 1)

FR<em>z </em>= (1/2) M*(v₂*Sin ∅) + (1/2) M*(-v₂*Sin ∅) = 0

Substituting the given values,

FR<em>x </em>= (5865.6 lbm/s)*(18 ft/s)*(Cos (45°) - 1)(1 lbf / 32.2 lbm*ft/s²)

⇒ FR<em>x </em>= - 960.37 lbf

FR<em>z </em>= 0 lbf

The negative value for FR<em>x</em> indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 960.37 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction. This can also be concluded from the symmetry.

In reality, the gravitational effects will cause the upper stream to slow down and the lower stream to speed up after the split. But for short distances, these effects are negligible.

3 0

2 years ago

At which lunar phase(s are tides least pronounced (e.g., the lowest high tides? at which lunar phase(s are tides least pronounce

Delicious77 [7]

Both first and third quarters. (:

7 0

2 years ago

1. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion begins to chase a fish at this speed when the fish is

Pie

#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

$d_1 = d_2 + L$

here

d1 = distance covered by sea lion in time t

d2 = distance covered by fish in the same time t

L = distance between fish and sea lion initially = 60 m

$d_1 = v_1 * t$

$d_1 = (40*\frac{5}{18})*t = \frac{100}{9}*t$

$d_2 = (16*\frac{5}{18})*t = \frac{40}{9}*t$

$\frac{100}{9}*t = \frac{40}{9}*t + 60$

$\frac{100}{9}*t - \frac{40}{9}*t = 60$

$\frac{60}{9}*t = 60$

$t = 9 s$

So it will take 9 s to chase the fish by sea lion

# 2

velocity of truck on road = 25 m/s along North

velocity of dog inside the truck = 1.75 m/s at 35 degree East of North

$v_{dt} = 1.75 cos35\hat j + 1.75sin35 \hat i$

$v_{dt} = 1.43 \hat j + 1 \hat i$

we can write the relative velocity as

$v_d - v_t = 1.43 \hat j + 1 \hat i$

$v_d = v_t + (1.43 \hat j + 1 \hat i)$

now plug in the velocity of truck in this

$v_d = 25 \hat j + (1.43 \hat j + 1 \hat i)$

$v_d = 26.43 \hat j + 1 \hat i$

so it is given as

$v_d = \sqrt{26.43^2 + 1^2} = 26.44 m/s$

direction will be given as

$\theta = tan^{-1}\frac{v_x}{v_y}$

$\theta = tan^{-1}\frac{1}{26.43} = 2.2 degree$

so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North

8 0

2 years ago

Read 2 more answers

A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. What is the acceleration of the ball in the vertical

blagie [28]

The acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Explanation:

This is a typical example of projectile motion, which consists of two independent motions:

- A uniform horizontal motion at constant velocity (since there are no forces in the horizontal direction)

- A vertical accelerated motion at constant acceleration (due to the presence of the force of gravity)

We are considering now the vertical motion only. There is only one force acting on the ball in this direction: the force of gravity, of magnitude

F = mg

where m is the mass of the ball and $g=9.8 m/s^2$ the acceleration due to gravity, downward. This means that the acceleration of the ball in this direction is (using Newton's second law)

$a=\frac{F}{m}=\frac{mg}{m}=g = 9.8 m/s^2$

Therefore, the acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

2 years ago