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3 years ago

What force is necessary to accelerate 32kg object by 2 m/s

Physics

1 answer:

Oxana [17]3 years ago

6 0

Answer:

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Explanation:

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The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ = 0.23805 N-m

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0

3 years ago

I need to know how to solve this problem for an upcoming exam

Nataly_w [17]

W=f*d = 5000N * 10 stories * each 4m = 200 000 J

7 0

3 years ago

a dock worker pushes a 150.0 kg crate up a ramp that is 3.00 m high and 10.00 m long into the cargo bay of a ship. He exerts a 8

juin [17]

Efficiency: (2940 J / 3250 J) x 100% = 90.46%

3 0

3 years ago

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. Th

jonny [76]

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

6 0

4 years ago