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bogdanovich [222]
2 years ago
8

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r

amp which ends 10 feet above a level landing zone. Assume the cyclist maintains a constant speed up the ramp and the ramp is inclined Ao (degrees) above horizontal. With the pictured imposed coordinate system, the parametric equations of the cyclist will be: x(t) = 100t cos(A) y(t) = –16t2 + 100t sin(A) + 10. (These are the parametric equations for the motion of the stunt cyclist.)If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A=
Physics
1 answer:
Elena-2011 [213]2 years ago
8 0

Answer:A=23.57^{\circ}

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of x(t)=100 t\cos (A)

y(t)=-16t^2+100t\sin (A)+10

Maximum height=35 feet

at maximum height

\frac{\mathrm{d} Y}{\mathrm{d} t}=0

\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A

t=3.125\sin A

At t=3.125\sin A

Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10

312.5\sin^2 A-156.25\sin^2 A=25

sin^2 A=0.16

A=23.57^{\circ}

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An electret is similar to a magnet, but rather than being permanently magnetized, it has a permanent electric dipole moment. Sup
steposvetlana [31]

Answer:

 F = 9.216 × 10⁻³ N

Explanation:

given,

dipole moment = 1 × 10⁻⁷ Cm

distance apart from +80 nC charge = 25 cm

to calculate the magnitude of electric force

Electric field due to dipole

E = \dfrac{2p}{4\pi \varepsilon_0 r^3 }

E = \dfrac{9\times 10^9\times 2\times 1 \times 10^{-7}}{25^3\times 10^{-6} }

E = 1.152 × 10⁵ N/C

electric force on the ball

F = E q

  = 1.152 × 10⁵ × 80 × 10⁻⁹

 F = 9.216 × 10⁻³ N  

Hence, the electric force is equal to  F = 9.216 × 10⁻³ N

7 0
2 years ago
Read 2 more answers
Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink?
bearhunter [10]

Answer:

It will sink

Explanation:

An object in the water can float only if its density is lower than the density of the water.

In fact, for an object completely immersed in water, there are two forces acting on it:

- Its weight, W=mg=\rho_o V g, downward, where \rho_o is the density of the object, V its volume and g the gravitational acceleration

- The buoyant force, B=\rho_w V g, upwards, there \rho_w is the density of the water

We see that when the density of an object is larger than the density of the water, \rho_o > \rho_w, the weight is greater than the buoyant force, W>B, so the object sinks.

In this case, the rock has a density of 1.73 g/cm3, while water has a density of 1.0 g/cm^3, so the rock will sink.


5 0
2 years ago
Read 2 more answers
Which of the following statements about stages of nuclear burning (i.e., first-stage hydrogen burning, second-stage helium burni
Darya [45]

Answer:

<em>C) Each successive stage lasts for approximately the same amount of time.</em>

<em></em>

Explanation:

Nuclear burning is a series of nuclear processes through which a star gets its energy. The energy within a star is due to nuclear fusion of lighter elements (hydrogen) into more massive element (helium), with a release of a large amount of energy due to the conversion of some of the mass into energy. Each stage leads to a loss of some of the mass which is converted into energy (option A is valid).

The fusion of four hydrogen atoms into one helium atom means that there is a creation of element with a higher atomic weight (option D is valid), and the energy output of each stage exceeds its energy input, meaning that each stage will require a higher temperature than its previous stages (option B is valid).

5 0
2 years ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft&gt;s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
2 years ago
Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
2 years ago
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