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valentina_108 [34]
3 years ago
10

Which is the mode of the following data?120, 125, 130, 125, 135​

Physics
2 answers:
Aleks [24]3 years ago
8 0

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

miss Akunina [59]3 years ago
5 0

Answer:

125

Explanation:

mode is most occurring number

in this data 125 comes two times other numbers only come once hence mode is 125

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Two friends leave a movie theater and take different busses to the same ice cream shop. One bus takes a longer route driving on
lisov135 [29]

Answer: short displacement has shorter road and long displacement has longer displacement . now think your self Which one is right

6 0
2 years ago
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
2 years ago
Power compression is caused by two things: 1) The heating up of a driver's voice coil which causes the impedance of the speaker
Morgarella [4.7K]

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7 0
3 years ago
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


5 0
2 years ago
A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc
max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }

f = \frac{1}{2\pi } ( 13.522)

f = 2.1535 Hz

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0
1 year ago
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