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Zinaida [17]
3 years ago
14

18. The displacement of an object moving 330 km North for 2 hours and an additional 220

Physics
1 answer:
motikmotik3 years ago
3 0

Answer:

Explanation:

1) Displacement of the object will be gotten using simply adding the distances since they are goinf in the same direction (positive y direction)

Displacement= 330km + 220km

Displacement = 550km North

2) For a falling body, the body will possess positive acceleration due to gravity because it is falling under the influence of gravitational force. If a body is not under the influence of gravity and is thrown up, such body wont come back down. Hence the value of acceleration due to gravity of a body falling to the ground is +9.8 m/s² (note that the unit of acceleration is m/s²)

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Which types of numbers does scientific notation best describe?
Travka [436]
The correct answer is
<span>c) very small and very large

Let's see this with a few examples:
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</span>0.0000000001
<span>we see that we can write it easily by using the scientific notation:
</span>1\cdot 10^{-10}
<span>2) Similarly, if we have a very large number:
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4 0
2 years ago
When you hold a frozen banana treat in your hands, which statement BEST describes what is taking place? A. Heat is flowing from
Andrej [43]

Answer:

The answer is A

Explanation:

It is A because your body heat is warmer than the banana and when you hold it the heat is transferring over.

3 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

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2 years ago
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