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3 years ago

What is the force weight of a jaguar who jumps 3 meters to a tree branch with 2670 J of work?

Physics

1 answer:

fgiga [73]3 years ago

7 0

Answer:

$we \: know \: energy \: = \: force \: \times distance \\ e = f \times d \\ so \: f \: = \frac{e}{d} \\ so \: th \: force \: here \: = (\frac{2670}{3}) newton \\ = 890newton$

Hope it helps

You might be interested in

An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at b

balandron [24]

Answer:

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

8 0

3 years ago

At what height from the surface of the earth does the value of acceleration due to gravity be 2.45 m/s square where the radius o

Citrus2011 [14]

Answer:

Explanation: RADIUS OF EARTH = 6400X1000m =

ACC DUE GRAVITY ABOVE SURFACE OF EARTH = g' =2.45 m/s^2

ACC DUE GRAVITY ON SURFACE OF EARTH =g= 9.8 m/s^2

A/C TO FORmULA

g'/g=1-2h/Re

g'/g +2h/Re = 1

2h/Re =1- g'/g

2h= (1- g'/g)Re

2h=(1-2.45 /9.8)

6400X1000

2h = (0.75)6400X1000

2h = 4800000

h= 2400000

3 0

2 years ago

A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere $I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

Explanation:

B = Bulk modulus

Intensity, $I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }$

The amplitude of oscillation of the sphere is given by:

$P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\$

$A = \triangle R\\$

Substitute v and A into Pmax

$P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}$

$I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }$

$P_{total} = 4\pi R^{2} I$

$P_{total} =4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }$

The intensity of the sound wave at a distance is given by:

$I = \frac{P_{total} }{4\pi d^{2} }$

$I = 4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

5 0

3 years ago

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)

Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

the x-component of the force at A is $A_{x}$ = 0

the y-component of the force at A is $A_{y}$ = 400 N

the couple acting at A is; $M_{A}$ = 146 N-m

the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑ $M_{L}$ = 0

Vector force acting at A, $F_{A}$ = $A_{x}i$ + $A_{y}j$

Now, From the image, we have;

∑f = 0

$F_{A}$ - 600j + 200j = 0i + 0j

$A_{x}i$ + $A_{y}j$ - 600j + 200j = 0i + 0j

$A_{x}i$ + ( $A_{y}$ - 400)j = 0i + 0j

now by equating i- coefficients'

$A_{x}$ = 0

so, the x-component of the force at A is $A_{x}$ = 0

also by equating j-coefficient

$A_{y}$ - 400 = 0

$A_{y}$ = 400 N

hence, the y-component of the force at A is $A_{y}$ = 400 N

we also have;

∑ $M_{L}$ = 0

$M_{A}$ - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

$M_{A}$ - 30 N-m - 228 N-m + 112 Nm = 0

$M_{A}$ - 146 N-m = 0

$M_{A}$ = 146 N-m

Therefore, the couple acting at A is; $M_{A}$ = 146 N-m

The sum of the moments about right end of the beam is;

∑ $M_{R}$ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)( $A_{y}$ ) + $M_{A}$

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑ $M_{R}$ = 0

Therefore, the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0

2 years ago

Why do we need the periodic table? First to answer get brainless!!!!!

mylen [45]

<span>The periodic table is the most important chemistry reference there is. It arranges all the known elements in an informative array. Elements are arranged left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass.

</span>

3 0

3 years ago

Read 2 more answers