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3 years ago

What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)

Physics

2 answers:

Lisa [10]3 years ago

8 0

Answer:

The answer would be 735J

Explanation:

PE=mgh

=(mass)(force of gravity)(height)

=(25kg)(9.8m/s^2)(3m)

=735J

tatiyna3 years ago

8 0

Explanation:

given, mass =25 kg

height=3m

and acceleration deu to gravity =9.8m/s^2

now, we have,

pe =mgh

=25×9.8×3

=735j.....ans

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A 2300 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow

IrinaK [193]

Answer:

a) $a_{max} = 3.8\,\frac{m}{s^{2}}$ , b) $F = 9260\,N$

Explanation:

a) The maximum possible acceleration that the truck can give to the SUV is:

$a_{max} = \frac{18,000\,N}{2,400\,kg+2300\,kg}$

$a_{max} = 3.8\,\frac{m}{s^{2}}$

b) The equation of equilibrium for the truck is:

$18,000\,N - F = (2,300\,kg)\cdot (3.8\,\frac{m}{s^{2}})$

The force of the SUV's bumper on the truck's bumper is:

$F = 9260\,N$

3 0

3 years ago

Read 2 more answers

The potential difference across a resistor in a particular electric circuit is 420 V. The current through the resistor is 19.0 A

damaskus [11]

Answer: 22.11 ohms

Explanation:

Potential difference(v)=420V

Current(I)=19.0 amp

Resistance(r)=?

r=v/I

r=420/19.0

r=22.11 ohms

7 0

3 years ago

The electric field at the center of a ring of charge is zero. At very large distances from the center of the ring along the ring

Daniel [21]

Answer:

z = 1.16m

Explanation:

The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:

$E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}$

z: distance to the plane of the ring

r: radius of the ring

Q: charge of the ring

you have that:

E_{z->0} = 0

E_{z->∞} = 0

To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:

$\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}\\\\$