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3 years ago

9

The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar

s? 41 N 5.0 N 19 N 8.4 N

1 answer:

allsm [11]3 years ago

6 0

Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N

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Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2

Mila [183]

Answer:

The angular acceleration is 11.66 rad/s²

Explanation:

Step 1: Given data

Three forces are applied to a solid cylinder of mass 12 kg

F1 = 15 N

F2 = 24 N

F3 = 19 N

R2 = 0.22m

R3 = 0.10m

Step 2: Find the magnitude of the angular acceleration

I = ½mr² = ½ * 12kg * (0.22m)² = 0.29 kg*m²

torque τ = I*α

τ = F2*R2 - F1*R1 = 24N*0.22m - 19N*0.10m = 3.38 N*m

This means

I = ½mr² = 0.29 kg*m²

τ = I*α = 3.38 N*m

OR

0.29 kg*m² * α = 3.38 N*m

α = 11.655 rad/s² ≈11.66 rad/s²

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3 years ago

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lesantik [10]

Answer:

240 meter

Explanation:

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3 years ago

A uniform plane wave traveling in air is incident upon a flat, lossless, and infinite in extent dielectric interface with a diel

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3 years ago

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

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To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

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h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

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Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

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