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3 years ago

An automobile with a mass of 1240.00 kg has 3.99 m between

Physics

1 answer:

Vitek1552 [10]3 years ago

8 0

Answer:

$948.15248\ N$

$5134.04751\ N$

Explanation:

$F_f$ = Force on front wheels

$F_r$ = Force on rear wheels

Distance between CG and rear wheel = 3.99-0.622 = 3.368 m

$F_r=1240\times 9.81-F_f$

As the forces are conserved we have

$(1240\times 9.81-F_f)3.368=F_f\times 0.622\\\Rightarrow 12164.4-F_f=\dfrac{0.622F_f}{3.368}\\\Rightarrow 12164.4-F_f=0.18467F_f\\\Rightarrow F_f=\dfrac{12164.4}{1.18467}\\\Rightarrow F_f=10268.09503\ N$

On rear wheels

$F_r=1240\times 9.81-10268.09503\\\Rightarrow F_r=1896.30497\ N$

The force on each rear whees is $\dfrac{1896.30497}{2}=948.15248\ N$

Force on each front wheel is $\dfrac{10268.09503}{2}=5134.04751\ N$

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

The work required is -515,872.5 J

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3 years ago

If 270 wats of power is used in 42 seconds how much work is done ?

pentagon [3]

Answer:

power is given by work done divide time taken

P=w/t

therefore Pt=w

270x42=11,340.J

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4 years ago

Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in diffe

koban [17]

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

$v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}\\$

but the time just before the balls hit the ground is

$t=\frac{2v_{0}sin\alpha}{g}$

Hence we have

ball A

$v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball B

$v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball C

$v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$