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dimulka [17.4K]
3 years ago
8

A car started from a rest and accelerated at 9.54 m/s2 for 6.5 seconds. How much distance was covered by the car?

Physics
1 answer:
ElenaW [278]3 years ago
5 0

Explanation:

S =ut + 1/2at^2

S = 0×6.5 + (1/2 × 9.54) × 6.5^2

S =0 + 4.77 ×42.25

S=201.5m

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An unknown fluid has a specific gravity of 0.750. What is the volume of 22.5 kg of this fluid?
Andru [333]
<h2>Option C is the correct answer.</h2>

Explanation:

Specific gravity of fluid = 0.750

Density of fluid = Specific gravity of fluid x Density of water

Density of fluid = 0.750 x 1000

Density of fluid = 750 kg/m³

Mass of fluid = 22.5 kg

We have

         Mass = Volume x Density

         22.5 = Volume x 750

         Volume = 0.03 m³ = 30 L

Option C is the correct answer.

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3 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

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using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

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A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

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Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

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