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Rudiy27
2 years ago
12

A toy car that 0.12 m long is used to model the actions of an actual car that is 6m long. Which ratio shows the relationship bet

ween the sizes of the model and the actual car?
Physics
1 answer:
ra1l [238]2 years ago
3 0

Answer:

We know that a 0.12m long toy car is used as a model of a 6m long car.

We want to find the ratio that shows the relationship between both cars (the toy one and the real one)

The ratio that shows the relationship between both measures is just the quotient between both lengths, for example, if we take:

0.12m/6m = 0.02

This means that the length of the toy car is 0.02 times the length of the long car.

We also could have:

6m/0.12m = 50

This means that the length of the real car is 50 times the length of the toy car.

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1. Given an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?
DanielleElmas [232]
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2 years ago
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The maximum speed limit on interstate 10 is 75 miles per hour. how many meters per second is this
Dvinal [7]

Answer:

<h2>33.53m/s</h2>

Explanation:

Given the maximum speed limit on interstate 10 as 75 miles per hour, to get the speed in meter per seconds, we need to convert the given speed to meter per seconds.

Using the conversion 1 mile = 1609.34m and 1 hour = 3600 seconds

75 miles perhour = 75miles/1 hour

75miles/1 hour (in m/s) = 75miles*1609.34m* 1 hour/1mile * 1 hour * 3600s *

= 75 *1609.34m* 1 /1 * 1 * 3600s

= 120,700.5m/3600s

= 33.53m/s

<em>Hence the maximum speed limit on interstate 10 in metre per seconds is 33.53m/s</em>

8 0
3 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi
iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

3 0
3 years ago
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