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Rasek [7]
3 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]3 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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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
3 years ago
At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A
saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B=0.6\times 10^{-4} T

\theta=75^{\circ}

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F=IBlsin\theta

Using the formula

F=0.6\times 10^{-4}\times 15\times 19sin75

F=16.5\times 10^{-3} N

Direction=tan\theta=cot(90-75)=tan15^{\circ}

\theta=15^{\circ}

15 degree above the horizontal  in the northward direction.

5 0
3 years ago
What H & N can friction produce?
Angelina_Jolie [31]
Friction produces heat hope this helps

5 0
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An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon
erastova [34]

The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ_{1} = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ_{2} = Adjacent side 2 / Hypotenuse

cos θ_{2} = 4.98 / 8.5 = 0.59

θ_{2} = cos^{-1} ( 0.59 )

θ_{2} = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

To know more about trigonometric ratios

brainly.com/question/1201366

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3 0
1 year ago
NEED HELP ASAP PLEASE!
katrin [286]

Answer:

Here is the solution hope it helps:)

5 0
2 years ago
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