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slamgirl [31]
3 years ago
13

A cleaver physics professor wants to create a situation where a block starts from rest at the top of a 31.0° inclined plane and

encounters a spring at the bottom of the incline. The spring has a constant 3.4 kN/m and the block's mass is 33.0 kg. How far does the block travel before hitting the spring, if the spring was compressed 37 cm in it's initial collision?
Physics
1 answer:
UNO [17]3 years ago
8 0

Answer:

Explanation:

Let the length of inclined plane be L .

work done by gravity on the block

= force x length of path

= mg sinθ x L , m is mass of the block , θ is inclination of path

This in converted into potential energy of compressed spring

1/2 k x² = mgL sin31  , k is force constant . x is compression

.5 x 3400 x .37² = 33 x9.8 x sin31 L

L = 1.4

Length of incline = 1.4 m .

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A student stands on a scale, and the scale reads 85 N. What is being measured by the scale
Dmitry [639]

Force. Force is measured by newtons, probably because he came up with the equations to calculate force.

Hope this helps!

5 0
3 years ago
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle
liq [111]

Answer:

0.0031792338 rad/s

Explanation:

\theta = Angle of elevation

y = Height of balloon

Using trigonometry

tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta

Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

6 0
3 years ago
MULTIPLE CHOICE QUESTION
Nookie1986 [14]

Which object? More information is needed to answer this question

4 0
3 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

5 0
3 years ago
Find the length of the third side of the triangle.
AlladinOne [14]

Answer:

v

Image result for What is the length of the hypotenuse of the triangle below

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.Nov 23, 2016

Explanation:

4 0
3 years ago
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