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Rasek [7]
2 years ago
10

Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one

is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Physics
1 answer:
Dvinal [7]2 years ago
6 0

Answer:

e.

Explanation:

  • Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according  Newton's 2nd Law, as we can see below:
  • F_{g}  = m*g = m*a  (1)
  • ⇒a = g = 9.8m/s² (pointing downward)
  • Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:

       \Delta y = v_{o} * t - \frac{1}{2} *g*t^{2}  (2)

  • Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:

        h =\frac{1}{2} *g*t^{2}  (3)

  • Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.
  • If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.
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A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2
butalik [34]

Answer:        

Number of turns in secondary will be 7

Explanation:

We have given primary voltage V_p=120volt

Number of turns in the primary is N_p=3575

Secondary voltage is given V_s=235mV=0.235volt

We have to find the number of turns in secondary

We know that \frac{N_p}{N_s}=\frac{V_p}{V_s}

So \frac{3575}{N_s}=\frac{120}{0.235}

N_s=6.60

As the number of turns can not be in decimal so number of turns will be 7

6 0
3 years ago
What type of heat transfer is used to fry an egg in a pan and how can you tell?
Deffense [45]

Answer: Option (B) is the correct answer.

Explanation:

A process in which transfer of heat occurring when one object touches another object is known as conduction.  

For example, when a pan is heated and then an egg is placed over the pan. Then there will be transfer of heat from pan towards the egg.

As a result, the egg gets cooked because the egg was touching the pan.

Hence, we can conclude that it is conduction because the egg is touching the pan.

6 0
2 years ago
Read 2 more answers
An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the
DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0
3 years ago
Why do scientist describe the properties of elemts under "Standard Conditions"?
skad [1K]
The standard states of elements are the forms that they adopt at a temperature of 25°C and pressure of 1 atmosphere. These forms of the elements are the reactants in the formation equations of multi-element substances. The heat of formation (∆Hf°) of an element in its standard state is zero
7 0
2 years ago
Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force
gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

F=k\frac{q_1q_2}{r^2}

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}

F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N

Ft=0N

Hope this helps!!

5 0
3 years ago
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