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Ksivusya [100]
3 years ago
7

A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the

object for the next 4.00 s. How far has the object moved during this 7.00 s interval?
Physics
1 answer:
olya-2409 [2.1K]3 years ago
7 0

Answer:

The total distance at 7 s is:

x_{tot}=27\: m

Explanation:

<u>Distance due to the force</u>

We can use second Newton's law to find the acceleration.

F=ma

a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

x_{1}=0.5at_{1}^{2}

x_{1}=0.5(1.8)(3)^{2}

x_{1}=8.1\: m

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s

So the second distance will be:

x_{2}=vt_{2}=5.4*4=21.6\: m

Therefore, the total distance is:

x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m

I hope it helps you!

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ryzh [129]

Answer:

The bond energy of F–F = 429 kJ/mol

Explanation:

Given:

The bond energy of H–H = 432 kJ/mol

The bond energy of H–F = 565 kJ/mol

The bond energy of F–F = ?

Given that the standard enthalpy of the reaction:

<u>H₂ (g) + F₂ (g) ⇒ 2HF (g)</u>

ΔH = –269 kJ/mol

So,

<u>ΔH = Bond energy of reactants - Bond energy of products.</u>

<u>–269 kJ/mol = [1. (H–H) + 1. (F–F)]  - [2. (H–F)]</u>

Applying the values as:

–269 kJ/mol = [1. (432 kJ/mol) + 1. (F–F)]  - [2. (565 kJ/mol)]

Solving for , The bond energy of F–F , we get:

<u>The bond energy of F–F = 429 kJ/mol</u>

5 0
4 years ago
A 2640-Hz sound source is moving at 15.0m/s toward a stationary observer. What is the frequency heard by the observer if the spe
Dvinal [7]

Answer:

The frequency heard by the observer is 2760.73 hertz.                 Explanation:

Frequency of source, f = 2640 Hz

Velocity of source, v_s=15\ m/s

The speed of sound, v = 343 m/s

Let f' is the frequency heard by the observer. According to Doppler's effect, the frequency of the observer is given by :

f'=\dfrac{fv}{v-v_s} (as the source is moving towards observer)

f'=\dfrac{2640\times 343}{343-15}  

f' = 2760.73 Hz

So, the frequency heard by the observer is 2760.73 hertz. Hence, this is the required solution.                              

8 0
4 years ago
How is the movement of planet 1 in this system influenced by the other two objects in this star system
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Answer:

gravity

Explanation:

if a Venus sized object is orbiting a star and a two neptune sized objects are orbiting closely the will pull and yank on the venus sized object heating it up and the star will pull back on the other panlets heating the insides of the planet (take io as an example)

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3 years ago
The magnetic field produced by a current in a long straight wire varies with respect to the radial distance r from the wire as G
Katarina [22]

Answer:

1/r

Explanation:

The magnetic field produced by a current in a wire decreases with distance from the wire, this means the magnetic field produced by a currents is inversely proportional to it's distance from the wire.

Let E be magnetic field and r distance of wire

We have that,

E is inversely proportional to r

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