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3 years ago

A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s Result

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densk [106]

The elevator is going up

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2 years ago

4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C

aniked [119]

Answer:

a) ΔL/L = F / (E A), b) $L_{f}$ = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

$L_{f}$ -L = L (F / EA)

$L_{f}$ = L + L (F / EA)

$L_{f}$ = L (1 + L (F / EA))

4 0

3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

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

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s

garik1379 [7]

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

4 0

3 years ago

Read 2 more answers

Write down the principal of lever

pashok25 [27]

Answer:

See the explanation below.

Explanation:

A lever is a simple machine that changes the magnitude and direction of the force applied to move an object. Minimizes the force needed to lift the object.

By means of the following image, we can see the principle of operation of a lever.

The load can be moved thanks to the force multiplied by the distance to the fulcrum.

3 0

3 years ago