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kicyunya [14]
2 years ago
5

Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of res

earch question o add another research question o use the very first source you find for your project O change the scope of your project​
Physics
1 answer:
Nesterboy [21]2 years ago
3 0

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

    This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

    Would adding another question to an already broad question help? No.

O Use the very first source you find for your project

    If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question

O Change the scope of your project​    

    You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p
Delicious77 [7]
<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
   m1 * v1 + m2 * v2 = m * v
 1 * 1 + 5 * 0 = (1 + 5) * v
  1 = 6 * v
 v = 1/6 m/s
   So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>
6 0
3 years ago
PLEASE HELP ME WITH THIS PROBLEM
valentinak56 [21]

1) The mass of the continent is 2.13\cdot 10^{21} kg

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

1)

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3

The mass of the continent is given by

m=\rho V

where:

\rho = 2790 kg/m^3 is its density

V=7.62\cdot 10^{17} m^3 is its volume

Substituting, we find the mass:

m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg

2)

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is

v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s

Now we can find the kinetic energy of the continent, which is given by

K=\frac{1}{2}mv^2

where

m=2.13\cdot 10^{21} kg is the mass

v=5.08\cdot 10^{-10}m/s is the speed

Substituting,

K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J

3)

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

K=\frac{1}{2}mv^2

where

v is the speed of the man

And solving the equation for v, we find his speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'
NNADVOKAT [17]

Answer:

Acceleration, a=-1.5\ m/s^2

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)  

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

v=u+at

a=\dfrac{v-u}{t}

a=\dfrac{-5-10}{10}    

a=-1.5\ m/s^2

So, the acceleration of the car is -1.5\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
3 years ago
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