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Klio2033 [76]
3 years ago
13

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

(a) 1.0 s, (b) 2.0 s, and (c) 3.0 s?
Physics
1 answer:
DENIUS [597]3 years ago
4 0

Explanation:

It is given that,

The acceleration of the toboggan, a=1.8\ m/s^2

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

s=ut+\dfrac{1}{2}at^2

At t = 1 s

s=\dfrac{1}{2}\times 1.8\times 1^2

s_1=0.9\ m

At t = 2 s

s=\dfrac{1}{2}\times 1.8\times 2^2

s_2=3.6\ m

At t = 3 s

s=\dfrac{1}{2}\times 1.8\times 3^2

s_3=8.1\ m

Hence, this is the required solution.

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I think option C is correct..hope it helps
3 0
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goldenfox [79]

Answer:

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Explanation:

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6 0
3 years ago
Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.
lapo4ka [179]

Answer:

This is how I do it:

  • 5.24 rounded to one significant figure is 5
  • 43.261 rounded to one significant figure is 40.
  • 7.3458 rounded to one significant figure is 7.
  • 5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0
3 years ago
The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia
MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

              How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

                    = 1 J - 0.313 J

                    = 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

                    = 1 J - 0.45 J

                    = 0.55 J

So,

Energy Lost for group B's car = 0.55 J

8 0
3 years ago
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma
Sindrei [870]

Answer:

E = 1.50 × 10^{8} V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

\mu _b = \frac{B}{2\mu _o}   ...............1

and

energy density due to electric filed is

\mu _e = \frac{\epsilon _o E^2}{2}     ...............2

and here \mu _b = \mu _ e

so that

E = \frac{B}{\sqrt{\mu _o \times \epsilon _o}}      ...................3

put here value and we get

E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}  

E = 3 × 10^{8}  × 0.50

E = 1.50 × 10^{8} V/m

6 0
3 years ago
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