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Elis [28]
2 years ago
11

A force of 16 N to the west is applied to each object below. Which object will

Physics
2 answers:
Montano1993 [528]2 years ago
7 0

41kg object that is moving east at 5 m s

kenny6666 [7]2 years ago
7 0

Answer:

41kg object that is moving east at 5 m s

Explanation:

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To ensure a steady flight, the standard golf ball has nearly 400 ____ or “dimples” on its surface.
Gekata [30.6K]
To ensure a steady flight, the standard golf ball has nearly 400 indentations <span>or “dimples” on its surface. The correct option among all the options that are given in the question is the second option or option "B". The other choices are incorrect. I hope that this is the answer that has actually come to your help.</span>
4 0
3 years ago
The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me
kondaur [170]

Answer with Explanation:

The force of attraction between 2 charges of magnitudeq_1,q_2 separated by a distance 'r' is given by

F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}

where

\epsilon _o is a constant known as permitivity of free space

k=9\times 10^{9}Nm^2/C^2

Applying the given values in the above relation we get

8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters

5 0
2 years ago
provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small
gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

T_{simple = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / I ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and I  is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

I = \frac{1}{3}mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/I) OR T = 2π√(I/mgL)

so we can use I = \frac{1}{3}mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L =  \frac{1}{2}D.

now, substituting these equations, the period becomes;

T = 2π/√(I/mgL) OR T = 2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } } OR T = 2π√(2D/3g )  ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω_{simple = 2π/T_{simple OR  ω_{simple = √(g/D) OR  ω_{simple = 2π√( D/g )  

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × T_{simple

we substitute in value of T_{simple

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

 

8 0
2 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
Below is a diagram of the orbits of the inner planets examine the diagram and answer questions three and four 
drek231 [11]

Answer:

C is the answer

Explanation:

Earth is the third planet from the sun

you are cheating on your test, I know

3 0
3 years ago
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