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Andrew [12]
3 years ago
7

Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là

30cm. Chiều dài nhỏ nhất và lớn nhất của lò xo trong quá trình dao động là bao nhiêu? Lấy g = 10m/s2.
Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

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Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram
bagirrra123 [75]

Speed of car A is given as

v_a = 70 mph

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

v_a = 70 *\frac{1609}{3600} = 31.3 m/s

now its distance from Bambi is given as

d_a = 350 m

time taken by it to hit the Bambi

t = \frac{d}{v}

t = \frac{350}{31.3}

t = 11.2 s

Now other car is moving at speed 50 mph

so its speed in SI unit will be

v_b = 50* \frac{1609}{3600}

v_b = 22.35 m/s

now its distance from Bambi is given as

d_b = 590 feet

as we know that 1 feet = 0.3048 m

d_b = 590*0.3048 = 179.83 m

now the time to hit the other car is

t_2 = \frac{179.83}{22.35}

t_b = 8.05 s

So Car B will hit the Bambi first

7 0
3 years ago
What happens when a proton is placed directly in the path of the proton cannon?
blondinia [14]

Answer:coherent light

Explanation:

5 0
3 years ago
If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

So you need about 30 RPM to keep people from falling out the bottom

7 0
4 years ago
a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the
irinina [24]
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


7 0
3 years ago
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0
3 years ago
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