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timofeeve [1]
3 years ago
14

Which type of lever is steep road​

Physics
1 answer:
Anni [7]3 years ago
3 0

Answer:

Third class lever

Explanation:

Third class lever

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Having greater biodiversity in an ecosystem is beneficial to humans because
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The answer is A

Explanation:

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Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
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2 years ago
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If a bullet travels at 593.0 m/s, what is its speed in miles per hour?
Ksenya-84 [330]

We have 1 \; mile = 1609.34\; meters. So, 1 \; meter = \frac{1}{1609.34} \;mile.

1 \; hour = 3600 \; s. So 1 \; s = \frac{1}{3600}  \; hour.

Thus we can convert the units of the given quantity.

That is,

593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour.

The quantity is converted to the required units.


7 0
3 years ago
Two children are balanced on a seesaw that has a mass of 18.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the
Mashcka [7]

Answer:

1.28 m

Explanation:

As shown in the diagram attached,

According to the principle of moment,

For a body at equilibrium,

Sum of clockwise moment = sum of anticlockwise moment.

Taking moment about the pivot,

W₁(1.6)+W(0.133) = W₂(x)............... Equation 1

Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.

But,

W = mg

Where g = 9.8 m/s², m = mass of the body

Therefore,

W₁ = 26×9.8 = 254.8 N,

Wₓ = 18×9.8 = 176.4 N

W₂ = 34.4×9.8 = 337.12 N

Substitute these values into equation 1

(254.8×1.6)+(176.4×0.133) = 337.12(x)

407.68+23.4612 = 337.12x

337.12x = 431.1412

x = 431.1412/337.12

x = 1.2789

x ≈ 1.28 m

7 0
2 years ago
You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the
podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

F_x = Fcos30


F_y = Fsin30


Now the normal force on the block is given as

N = Fsin30 + mg

N = 0.5F + (18\times 9.8)

N = 0.5F + 176.4

now the friction force on the cart is given as

F_f = \mu N

F_f = 0.625(0.5F + 176.4)

F_f = 110.25 + 0.3125F

now if cart moves with constant speed then net force on cart must be zero

so now we have

F_f + F_x = 0

Fcos30 - (110.25 + 0.3125F) = 0

0.866F - 0.3125F = 110.25

F = 199.2 N

so the force must be 199.2 N

8 0
3 years ago
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