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KATRIN_1 [288]
3 years ago
5

Numerical Problems:

Physics
1 answer:
dangina [55]3 years ago
6 0
  • Displacement=1200m
  • Time=4min=4(60)=240s

\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

\\ \sf\longmapsto Velocity=\dfrac{1200}{240}

\\ \sf\longmapsto Velocity=5m/s

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A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of th
AlekseyPX

0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)

35 / 1.2 = 29.16

29.16 ÷ 100 = 0.29

Wave velocity in string:

The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.

Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.

Learn more about density here:

brainly.com/question/15164682

#SPJ4

3 0
2 years ago
A dentist uses a concave mirror (focal length 2.5 cm) to examine some teeth. If the distance from the object to the mirror is 1.
AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

Thus the magnification of the tooth = 2.28.

3 0
3 years ago
You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0
2 years ago
A motorbike reaches a speed of 20 m/s over 60m, whilst
Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation  </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration  

An equation of uniformly accelerated motion  

V = vo + at  

Vt² = vo² + 2a (x-xo)  

x = distance on t  

vo / vi = initial speed  

vt / vf = speed on t / final speed  

a = acceleration  

vf=20 m/s

d = 60 m

a = 3 m/s²

\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s

7 0
3 years ago
A ball is thrown straight up at time t=0 with an initial speed of 19m/s. Take the point of release to be y0=0 and upwards to be
Wittaler [7]
First we write the corresponding kinematics equations:
 a = -g
 v = -g * t + vo
 y = -g * ((t ^ 2) / 2) + vo * t + yo
 Substituting the values:
 y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
 answer:
 the displacement at the time of 0.50s is 8.27m
4 0
3 years ago
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