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Nataly_w [17]
4 years ago
10

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

mass is about 1/9 and radius 1/2 that of Earth, is mostnearly
A) 1/3 T
B) 2/3T
C)T
D) 3/2T
E) 3 T
Physics
1 answer:
olchik [2.2K]4 years ago
6 0

Answer:

C)T

Explanation:

The period of a mass-spring system is:

T=2\pi\sqrt\frac{m}{k}

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

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A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causes the crate to travel at constant ve
koban [17]

Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!

tinyurl.com/wtjfavyw

4 0
3 years ago
In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation bet
Elina [12.6K]

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

7 0
4 years ago
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How long will it take a car to travel a distance of 1km if it has an average speed of 60kph​
beks73 [17]

Answer:

1÷60 h

time equals distance upon speed

6 0
3 years ago
A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t
Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height  is given by;

v = u - gt

u = gt

t = \frac{u}{g}

Time of flight, T = 2t

T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0
3 years ago
A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
3 years ago
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