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evablogger [386]
3 years ago
13

A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?

Physics
2 answers:
Brums [2.3K]3 years ago
7 0
It would have no “weight” since weight is determined by the force normal and since it is accelerating there is no normal force making it weightless
patriot [66]3 years ago
6 0
Objects in free fall are weightless.
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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y =  \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} }  =2\ m/s

Also:

\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x =  u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s

4 0
3 years ago
The density of aluminum is 2.7 g/cm3. A metal sample has a mass of 52.0 grams and a volume of 17.1 cubic centimeters. Could the
Fudgin [204]
 Answer:  
__________________________________________________
            No;  the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
________________________________________________
Explanation:
________________________________________________
Density is expressed as "mass per unit volume" ;

  in which:
     "mass, "m", is expressed in units of "g" (grams);  and:
     "Volume, "V", is expressed in units of "cm³ " (such as in this problem); or                                                   in units of "mL" ;
__________________________________________________
            {Note the exact conversion:  " 1 cm³ = 1 mL " .}. 
__________________________________________________
  The formula for density:  D = m/V ;

Given:  The density of aluminum is:  2.7 g/cm³.

Given:  A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
_________________________________________________________
Let us divide the mass of the sample by the volume of the sample;
by using the formula:
___________________________________________
            D = m / V ;  

     and see if the value is at, or very close to "2.7 g/cm³ ".  

If it is, then it could be aluminum.
____________________________________________________
The density for the sample:

  D = (52.0 / 17.1)   g/cm³ = 3.0409356725146199 g/cm³ ;
                                              →round to "3 significant figures" ;
                                          = 3.04 g/cm³ .
_______________________________________________
No; the sample could not be aluminum; since the density of aluminum, 
   "2.7 g/cm³ "   is NOT close enough to the density of the sample,
                        "3.04 g/cm³ " .
____________________________________________________
5 0
3 years ago
A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig
Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

4 0
2 years ago
Explain you could use a battery, wire and compass to
gladu [14]

Answer:

oh I'm so sorry I can't answer your question it has been a long time since I learned that. so I totally forgot how to do this. sorry!

5 0
3 years ago
Peering through a telescope could be considered Step 1 of the scientific method.<br> True<br> False
aleksklad [387]

Answer:

<em><u>True:</u></em> because when u look thru a telescope you are making an observation

Explanation:

4 0
2 years ago
Read 2 more answers
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