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Citrus2011 [14]
3 years ago
5

What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?​

Physics
2 answers:
REY [17]3 years ago
6 0

Answer:

450cm^2

Explanation:

Length (l) = 30cm

Width (w)= 15cm

Area = l × w = 30cm × 15cm = 450cm^2

Igoryamba3 years ago
5 0

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

You might be interested in
An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south
vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

8 0
3 years ago
A 120.0 kg crate is placed on a 15.00°
Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0
3 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0
2 years ago
While looking at a cliff, you observe that three visible layers of rocks are tilted about 30 degrees. There are four straight ho
xxTIMURxx [149]

Answer:

the principle of original horizontality and the principle of superposition

Explanation:

The <em>principle of horizontality</em> states that layers of sediment are originally deposited horizontally under the influence of gravity.

The <em>principle of superposition</em> states that the oldest layer layer is at the bottom and each layer above it is younger, with the youngest being at the top.

Unconformities help us find the age of different layers. An unconformity is a surface in which no new solid matter is deposited after a long geologic interval. <em>Angular unconformity </em>is a type of unconformity which different kinds of stratum were tilted or folded before deposition of younger layers of solid matter above the unconformity. Once the layers were folded and tilted, the older layers of the solid matter eroded, then the younger layers were deposited on the older layers. There <em>angular unconformity </em>is the contact between young and old layers of solid matter.

Therefore, these two principles therefore describe how the tilted layers are older than horizontal layers.

3 0
3 years ago
Consider a continuous time signal, x(t) = ∑ Ai 5 i=1 cos(2πFi t + θi), consisting of a sum of sinusoidal signals of frequencies
Wewaii [24]

Answer:

duhhhhh

Explanation:

duhhhhh

4 0
3 years ago
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