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Jobisdone [24]
3 years ago
8

A piece of metal is 4 cm by 10 cmby 2 cm. Find its mass if it has adensity of 2 g/cm'.​

Physics
1 answer:
harina [27]3 years ago
3 0

Answer:

m=ρV

V=LxWxD

V=4x10x2=80

m=2*80=160 grams

Explanation:

Mass is equal to density multiply by volume of object. Volume of rectangle pice can be calculated by multiplying all sides.

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A transformer is designed to convert the voltage from 240 V to 12 V, and has 4000 turns on its primary coil. The turns on its se
harkovskaia [24]

Answer:

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Explanation:

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4 0
3 years ago
A 5000 g toy car starts from rest and moves a distance of 300 cm in 3 s under the action of a single constant force. Determine t
sveticcg [70]

Answer:

3.33 N

Explanation:

First, find the acceleration.

Given:

Δx = 3 m

v₀ = 0 m/s

t = 3 s

Find: a

Δx = v₀ t + ½ at²

3 m = (0 m/s) (3 s) + ½ a (3 s)²

a = ⅔ m/s²

Use Newton's second law to find the force.

F = ma

F = (5 kg) (⅔ m/s²)

F ≈ 3.33 N

4 0
3 years ago
A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the
ehidna [41]

Answer: 170.67 N

Explanation:

Given

Mass of skier is m=82.9\ kg

Height of the inclination is h=20\ m

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N

Thus, the horizontal friction force is 170.67 N.

7 0
3 years ago
A squirrel drops an acorn from a tree. Starting from rest, it reaches the ground 22.0 meters
Llana [10]
Does it not tell you how long it took it to reach the ground? Constant Velocity should be distance over time
7 0
3 years ago
A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

7 0
3 years ago
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