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krek1111 [17]
3 years ago
9

When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If

the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?

Physics
1 answer:
Alex787 [66]3 years ago
6 0

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

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Find acceleration given the following:
Lorico [155]

Answer:

Explanation:df=378

Di=0

Vi=0

T=6s

Formula

Df=Di+Vit+1/2(at)2

378=0+0+6+1/2×a×t2

378=6+1/2×a×36

378=6+18a

18a=378+6=384

a=385/18=21.33(cm/s)/s

5 0
3 years ago
If I weighed 130 pounds what would my mass be in kilograms
lawyer [7]
You would weigh 58.967 kilograms
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Calculate the density of water in kg/m3. Express your answer in kilograms per cubic meter using three significant figures.
Svet_ta [14]

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, the density of water is generally known to be 997 kg/m³. The formula used to determine density is mass ÷ volume. In the case of the density of water, the mass is measured in kilograms (kg) while the volume is measured in cubic meter (m³).

8 0
3 years ago
Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N  , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N ,  θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y=  -46.1 *sin28.8° =  -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N  -22.2 N =  + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey=  - 8.4 N

Magnitude of the equilibrant (Fe)

F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2}  }

F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2}  }

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )

\beta = tan^{-1} (\frac-8.4 }{26.92 } )

β = -18.34°                  

β = 18.34° , clockwise from the positive x axis

8 0
3 years ago
A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0
otez555 [7]
A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
x(t)=v_0 \cos \alpha t
y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2
where
- the horizontal motion is a uniform motion, with constant speed v_0 \cos \alpha, where v_0 = 8.00 m/s and \alpha=20.0^{\circ}
- the vertical motion is an uniformly accelerated motion, with constant acceleration g=9.81 m/s^2, initial position h (the height of the building) and initial vertical velocity v_0 \sin \alpha (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
0=h-v_0 \sin \alpha t -  \frac{1}{2}gt^2
which becomes
h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that 
y(t)=h-10
If we substitute this into the equation of y(t), we have
h-10 = h-v_0 \sin \alpha t-  \frac{1}{2}gt^2
\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0
4.9 t^2 +2.74 t-10 =0
whose solution is t=1.18 s (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0
4 years ago
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