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3 years ago

A pilot drops a bomb from a plane flying horizontally. Where will the plane be located when the bomb hits the ground

Physics

1 answer:

kumpel [21]3 years ago

3 0

Answer:

The plane will be located directly above the bomb because they both have the same horizontal speed.

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A car starts from rest at the top of a hill with 45 J of gravitational

kherson [118]

Answer:

The car will be moving at 5.48 m/s at the bottom of the hill

Explanation:

Principle of Conservation of Mechanical Energy

In the absence of friction, the total mechanical energy is conserved. That means that

$E_m=U+K$ is constant, being U the potential energy and K the kinetic energy

U=mgh

$\displaystyle K=\frac{mv^2}{2}$

When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:

$\displaystyle \frac{mv^2}{2}=45$

Multiplying by 2:

$\displaystyle mv^2=90$

Dividing by m:

$\displaystyle v^2=\frac{90}{m}$

Taking square roots:

$\displaystyle v=\sqrt{\frac{90}{m}}$

$\displaystyle v=\sqrt{\frac{90}{3}}$

$v=\sqrt{30}$

v = 5.48 m/s

The car will be moving at 5.48 m/s at the bottom of the hill

3 0

3 years ago

At a distance of 0.220 cm from the axis of a very long charged conducting cylinder with radius 0.100cm, the electric field is 49

Scorpion4ik [409]

Answer:

At the distance of 0.220cm from the axis.

r = 0.220cm = 0.0022m, E = 490N/C, e0 = 8.854 x 10^-12F/m

Linear charge density = 2*π*e0*r*E = 2 x 3.142 x 8.854x10^-12 x 0.0022 x 490 = 5.998 x 10^-11C/m

Thus, To Calculate the Electric field at the distance r = 0.616cm from the cylinder axis, we substitute the calculated linear change density in the equation

E = (linear charge density)/2*π*e0*r

Here, r = 0.616cm = 0.00616m

E = [(5.998 x 10^-11)/(2 x 3.142 x 8.854 x 10-12 x 0.00616)]

E = 175N/C

Explanation:

The Electric field of a charged conducting cylinder obey the Gauss Law.

Therefore, the Electric field is given as:

E = (linear charge density)/4πe0r,

Where e0 is the permittivity of free space with constant value of 8.854 x 10^-12F/m, r is the radial distance from the axis.

3 0

3 years ago

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

$qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s$

So, the speed of the electron is 2591.83 m/s.

8 0

3 years ago

Why do faults often occur along plate boundaries?

Slav-nsk [51]

Earthquakes can also occur far from the edges of plates, along faults. Faults are cracks in the earth where sections of a plate (or two plates) are moving in different directions. Faults are caused by all that bumping and sliding the plates do. They are more common near the edges of the plates.

4 0

3 years ago

Becomes solid above the melting point

Whitepunk [10]

All liquids become solid, above the melting point!!! Hope this helps!!!

5 0

3 years ago