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Anit [1.1K]
2 years ago
3

Amir starts riding his bike up a 200-m-long slope at a speed of 18 km/h, decelerating at 0.20 m/s2 as he goes up. at the same in

stant, becky starts down from the top at a speed of 6.0 km/h, accelerating at 0.40 m/s2 as she goes down. how far has amir ridden when they pass?
Physics
1 answer:
BARSIC [14]2 years ago
4 0
The first thing we must know for this case are the following unit conversions:
 1 km = 1000 m
 1 h = 3600 s
 Now we pass the speeds to m / s:
 v1 = 18 * 1000/3600 = 5m / s
 v2 = 6 * 1000/3600 = 1.66 m / s
 Then, we write the kinematic formulas that describe the problem:
 d = v0 * t + (1/2) * a * t ^ 2
 For Amir,
 d1 = 5 * t - (1/2) * (0.20) * t ^ 2
 For Becky,
 d2 = (1.66) * t + (1/2) * (0.40) * t ^ 2
 On the other hand:
 d1 + d2 = 200
 Adding both equations we have:
 6.66 * t + 0.1 * t ^ 2 = 200
 Rewriting:
 0.1 * t ^ 2 + 6.66 * t - 200 = 0
 Solving the polynomial:
 t = 22.46 s
 Then, for Amir we have:
 d1 = 5 * t - (1/2) * (0.20) * t ^ 2
 d1 = 5 * (22.46) - (1/2) * (0.20) * (22.46) ^ 2
 d1 = 61.85m
 answer:
  Amir has ridden 61.85m when they pass
Guest
1 year ago
Followed this method and got the right answer for my problem. Thank you.
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Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

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2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

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2 years ago
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4 0
2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
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Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

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2 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
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Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

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V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

8 0
2 years ago
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