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avanturin [10]
2 years ago
11

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

which the wrecking ball strikes the building. The wrecking ball strikes the building with a force of N.
Physics
1 answer:
Y_Kistochka [10]2 years ago
3 0

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

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To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change
Yuki888 [10]

Answer:

Δ P =  13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂  = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P =  ρ₁ g Δh

Now by putting the values

Δ P =  9 x 10² x 9.81 x 1.5 x 10⁻³    Pa

Δ P =  13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.  

       

3 0
2 years ago
Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
2 years ago
The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec
saveliy_v [14]

The static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta.

The given parameters;

  • <em>mass of the block, = M</em>
  • <em>coefficient of static friction in section 1, = </em>\mu_s_1<em />
  • <em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta

Thus, the static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta

Learn more here:brainly.com/question/17237604

3 0
2 years ago
When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are
Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

5 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
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