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kolbaska11 [484]
3 years ago
13

A 5.0 kg mass is suspended from a spring. Pulling the mass down by an additional 10 cm takes a force of 20 N. If the mass is the

n released, it will rise up and then come back down. How long will it take for the mass to return to its starting point 10 cm below its equilibrium position?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

0.99 seconds

Explanation:

The problem depicts a simple harmonic motion.

Now, from Hooke's law,

The spring constant is given as;

k =F/x

where:

x is the displacement of the spring's end from its equilibrium position

F is the restoring force exerted by the spring on that end

From the question, F = 20N while x = 10cm = 0.1m. Thus,

K = 20/0.1 = 200 N/m

Now, time to take for the mass to return to its starting point is a period

The period oscillation of the mass is given as;

T = 2π√m/k

Where m = mass = 5kg

T = 2π√(5/200)

T = 0.99 s

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What are the two ways to create plasma?
dimaraw [331]

Answer:

Plasma can be artificially generated by heating a neutral gas or subjecting it to a strong electromagnetic field to the point where an ionized gaseous substance becomes increasingly electrically conductive.

5 0
3 years ago
Read 2 more answers
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0
2 years ago
a constant force acts on an object of mass 10 kg for a duration of 2 seconds it increases the object velocity from 4metre per se
inn [45]

Answer:

F=ma

a=v/t

a=vf-vi/t

a=8-4/2

a=4÷2

a=2msec*2

F=10×2

F=20N

3 0
3 years ago
A boy is standing motionless on a skateboard. He throws a basketball forward. Describe his motion.
weqwewe [10]

Answer:

Random motion

Explanation:

If the boy throws the basketball forward while at a position on the skateboard, the motion of the ball will be a random motion since we are not told if the ball is moving on a straight line when thrown forward.

In this case, the boy will tend to move in the direction of the ball. Since the ball is moving in a random manner, the motion of the boy will also be a random motion.

A random motion is a motion of a body in a zig zag manner. It is also known as Brownian motion e.g motion of a buzzing mosquito, motion of a smoke coming out of a chimney etc.

8 0
3 years ago
A thin beam of light of wavelength 50 μm (in the infrared portion of the em spectrum) goes through
Brut [27]

Solution :

Given :

Wavelength of the thin beam of light, λ = 50 μm

Distance of the screen from the slit, D = 3.00 m

Width of the fringe, Δy = ±8.24 mm

Therefore, width of the slit is given by :

$d=\frac{n \lambda D}{\Delta y}$

  $=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$

  = 0.000018203 m

  = 0.0182 mm

  = 0.018 mm

The intensity of light is given by :

$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$   , where $\beta=\frac{2 \pi D \sin \theta}{\lambda}$

$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

               = 0.1854

               ≈ 0.18

$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$

  $=I_0 \times 0.81$

  = 2  x0.81

  $= 1.62 \ W/m^2$

 

6 0
3 years ago
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