1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Aleks [24]
3 years ago
5

A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from its left end. If a force of 50N is appl

ied on it's right end, how much force would need to be applied to the left end?
100N
25N
200N
50N
​
Physics
1 answer:
grigory [225]3 years ago
6 0
The force that would be apply is
100N
You might be interested in
A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
A 0.60 kg rubber ball has a speed of 2.0 m/s at point A, and kinetic energy of 7.5 J at point
aliina [53]
<span>Let's first off calculate the kinetic energy using the formula 1/2MV^2. Where the mass, M, is 0.6Kg. And speed, V, is 2. Hence we have 1/2 * 0.6 * 2^2 = 1.2J. Since kinetic energy is energy due to motion; hence at point B the rubber has a KE of 1.2J and not 7.5J. So I would say that only the Mass and speed is actually true; While it's kinetic energy is not true.</span>
7 0
3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
Using Ohm’s Law, explain how resistance changes in relation to current, assuming that voltage remains constant.
Cerrena [4.2K]
V=IR so voltage is directly proportional to current. So for a given resistance increasing the voltage will result in a high current as well. This is because resistance is proportional to the voltage over the current. Ex: I=V/R
Hope this helped. THANKYOU for asking. <span />
7 0
3 years ago
On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou
Vilka [71]
The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2
3 0
3 years ago
Other questions:
  • Helpppp me solve this
    12·2 answers
  • if an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?
    5·2 answers
  • What is the difference between clastic and bioclast?
    12·1 answer
  • Biker A is cruising at a speed of 10.0 m/s when she passes biker B who is at rest at the origin of the coordinate system. Biker
    14·1 answer
  • Place these bodies of our solar system in the proper order of formation
    5·2 answers
  • What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
    9·1 answer
  • When you stand outside your are standing in the, Stratosphere. Troposphere. Mesosphere no
    12·1 answer
  • A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the gir
    12·1 answer
  • The Celsius temperature of –273° C is termed “absolute zero” and is the initial value on the metric unit of temperature, the Kel
    5·1 answer
  • A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!