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GREYUIT [131]
3 years ago
14

A conveyer belt moves a 40 kg at a velocity of 2/ms what is the kinetic energy of the box while it is on the conveyor belt

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0
KE = (1/2)·(mass)·(speed)²

It doesn't matter whether the object is in a car, on a boat, 
falling, on a conveyor belt, or being carried by ants.

KE = (1/2)·(40 kg)·(2 m/s)²

KE = (20 kg)·(4 m²/s²)

KE = 80 kg-m²/s²

KE = 80 Joules
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Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
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Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

\underset{V_{B}}{\rightarrow} = velocity of car B before collision = V_{B} i + 0 j

\underset{V_{AB}}{\rightarrow} = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

M_{A} = mass of car A = 1750 kg

M_{B} = mass of car B = 1450 kg

Using conservation of momentum

M_{A}  \underset{V_{A}}{\rightarrow} + M_{B}  \underset{V_{B}}{\rightarrow} = (M_{A} + M_{B}) ( \underset{V_{AB}}{\rightarrow} )

(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

V_{B} = 67.3 km/h        and  V_{A} = 34.4 km/



6 0
4 years ago
This will blow u r brains
ikadub [295]

Answer:

The last is proven

Explanation:

The moon landing is real and if you watch a video with Adam ruins everything then he debunks the theory

4 0
3 years ago
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The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area
postnew [5]

Answer:

Explanation:

Given that,

Small piston Hydraulic piston has an area

A1 = 0.01m²

If the force applied is 250N is applied to the small piston at an area of 0.05 m²

Then,

F2 = 250 N and A2 = 0.05m²

Then, applying pascal principle,

Pressure at small area = pressure are bigger area

P1 = P2

F1 / A1 = F2 / A2

F1 / 0.01 = 250 / 0.05

F1 / 0.01 = 5000

Cross multiply

F1 = 5000 × 0.01

F1 = 50 N

4 0
3 years ago
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A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length
Ad libitum [116K]

Answer:

a)  I = 0.0198 kg m² ,  b)    I = 21.85 kg m²

Explanation:

For this exercise we will use the definition of moment of inertia

        I = ∫ r² dm

For body with high symmetry they are tabulated

sphere  I = 2/5 m r²

bar with respect to  center of mass I = 1/12 m L²

let's calculate the mass of each body

bar

        ρ = m / V

        m = ρ V

        m = ρ l w h

where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm

        m = 32820 1 0.008 0.04

        m = 10.5 kg

Sphere

       M = ρ V

       V = 4/3 pi r³

       M = rgo 4/3 π r³

give us the density 37800 kg / m³ and the radius of 5 cm

       M = 37800 4/3 π 0.05³

       M = 19.8 kg

a) asks us for the moment of inertia of the sphere with respect to its center of mass

        I = 2/5 M r²

        I = 2/5 19.8 0.05²

        I = 0.0198 kg m²

b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes

        I = I_cm + M d2

where d is the distance from the body to the point of interest

        I_cm = 0.0198 kg m²

the distance to the pivot point is

        l = length of the bar + radius of the sphere

        l = 1 + 0.05 = 1.005 m

        I = 0.0198 + 19.8 1.05²

        I = 21.85 kg m²

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4 years ago
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