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3 years ago

How many forces are acting on a stationary raft floating in a swimming pool A1 B2 or C3

1 answer:

3 0

There are at least two forces on it, and there could be more.

Vertical forces:
-- gravity, directed downward
-- buoyant force, directed upward
These two forces must be exactly equal, so that the net
vertical force on the raft is zero. Otherwise, it would be
accelerating either up or down.

Horizontal forces:
We know that the net horizontal force on the raft is zero.
Otherwise, it would be accelerating horizontally.

But we don't know if there are actually no horizontal forces
at all, or a balanced group of horizontal forces, that add up
to a net force of zero.

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A car traveling in a constant speed of 55km/h on a circular track what is the acceleration explain

a_sh-v [17]

Answer:

See the explanation below

Explanation:

We must solve this problem by defining that when we have a constant velocity, the acceleration is equal to zero. That is, when there is no speed change, there is no acceleration. We can understand it very easily by means of the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 55 [km/h]

Vo = initial velocity = 55 [km/h]

a = acceleration [m/s²]

t = time [s]

As we can see there is no change in speed, and the difference between the two is equal to zero.

$0 = 0 +a*t\\a = (0-0)/t\\a= 0$

3 0

3 years ago

a 230-newton box rests on a plank 5 meters long. One end is 1.2 meters higher than the other end. Find the components of the for

NISA [10]

Answer:

55.3 N, 223.3 N

Explanation:

First of all, we can find the angle of the inclined plane.

We have:

L = 5 m the length of the incline

h = 1.2 m is the height

We also have the relationship

$h = L sin \theta$

where $\theta$ is the angle of the incline. Solving for the angle,

$\theta= sin^{-1} (\frac{h}{L})=sin^{-1} (\frac{1.2 m}{5 m})=13.9^{\circ}$

Now we can find the components of the weight of the box, which is the force that the box exerts on the plank. Calling W = 230 N the weight of the box, we have:

- Component parallel to the incline:

$W_{par} = W sin \theta = (230 N)(sin 13.9^{\circ}) =55.3 N$

- Component perpendicular to the incline:

$W_{per} = W cos \theta = (230 N)(cos 13.9^{\circ}) =223.3 N$

4 0

3 years ago

Ik this is a lot so 15 points and ill give brainliest thank you!! Questions should be very easy tho

Gnesinka [82]

1.plants,most algae and cyanobacteria

2.during photosynthesis cells use carbon dioxide and energy from the sun to make sugar molecules and oxygen

3.carbon dioxide+water---Glucose+oxygen

4.chloryphil

5.it absorbs blue and red good and absorbs green poorly

6 In plants, photosynthesis takes place in chloroplasts, which contain the chlorophyll. Chloroplasts are surrounded by a double membrane and contain a third inner membrane, called the thylakoid membrane, that forms long folds within the organelle

4 0

3 years ago

A ball is dropped from an initial height and allowed to bounce repeatedly. On the first bounce (one up-and-down motion), the bal

mojhsa [17]

Answer:

S₁₀ = 241.5837 m

Explanation:

h₁ = 2*32 in = 64 in

h₂ = 0.75*h₁ = (0.75)*64 in = 48 in

h₃ = 0.75*h₂ = 0.75*(0.75*h₁) = (0.75)²*h₁ = (0.75)²*64 in = 36 in

h₄ = 0.75*h₃ = 0.75*(0.75*h₁) = (0.75)³*h₁ = (0.75)³*64 in = 27 in

...

h₁₀ = 0.75*h₉ = (0.75)⁹*h₁ = (0.75)⁹*64 in = 4.8054 in

It is a geometric sequence (geometric progression) where the common ratio is

r = 0.75

Finding the sum of terms in a geometric progression is easily obtained by applying the formulas:

10<em>th</em> partial sum of a geometric sequence

S₁₀ = h₁*(1 - r¹⁰) / (1 - r)

⇒ S₁₀ = 64*(1 - 0.75¹⁰) / (1 - 0.75)

⇒ S₁₀ = 241.5837 m

6 0

4 years ago

An object whose mass is 300 lb experiences changes in its kinetic and potential energies owing to the action of a resultant forc

aev [14]

Answer: initial speed = 151.89 ft/s

Explanation:

1) About the question:

The question is incomplete. The final missing part is: If the object's elavation increases by 100 ft and its final velocirty is 200 ft/s, what is its initial velocity, in ft/s? Let g= 32.2 ft/s^2

2) Physical principle:

Work = change in mechanical energy

3) Formulas

Kinetic energy, KE = (1/2)mv²

Potential energy, PE = mgh

Mechanical energy, ME = KE + PE

4) Conversion of units:

i) 300 lbs = 300/2.2046 kg = 136.08 kg

ii) 140 btu = 140 × 1055.06 J = 147,708.4 J

iii) g = 9.81 m/s²

iv) 100 ft = 100 × 0.3048m = 30.48 m

v) 200 ft/s = 200 × 0.3048 m/s = 60.96 m/s

5) Solution

i) change in PE:

ΔPE = mgΔh = 136.08kg × 9.81 m/s² × 30.48m = 40,689.12 J

ii) change in KE

ΔKE = work done on the object - ΔPE = 147,708.4J - 40,689.12J = 107,019.28 J

iii) initial KE

initial KE = final KE - ΔKE = (1/2)mv² - ΔKE = (1/2) (136.08kg)(60.96m/s)² = 252,844.91J - 107,019.28J = 145,825.63J

iv) initial speed

KE = (1/2)mv² ⇒ v² = 2KE / m = 2 (145,825.63J) / 136.08 kg = 2,143.23 m²/s²

v = √v² = 46.30 m/s

v) Convert to ft/s

46.30 m/s = 46.30 / 0.3048 ft/s = 151.89 ft/s

7 0

4 years ago