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Jet001 [13]
2 years ago
15

The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic

h part of the journey was the car accelerating? c) Calculate the magnitude of average velocity of the car.

Physics
1 answer:
Gekata [30.6K]2 years ago
3 0

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is =\frac{0+40}{2\\}   {initial velocity + final velocity}

                                                            =\frac{40}{2}

                                                             =20

Therefore, the magnitude of the average velocity  of the car is 20 m/s

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wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
A group of students must conduct an experiment to determine how the location of an applied force on a classroom door affects the
schepotkina [342]

Answer:

the answer the correct one is the  d

Explanation:

In the gate rotation experiment several things are measured.

- the distance from the hinges to the applied force, which must be measured with a tape measure

- The value of the force that is devised with a dynamometer

- the rotated angle that is measured with a protractor

- the time it takes to turn an angle, which is measured with a stopwatch

When examining the answer the correct one is the  d

8 0
2 years ago
This is a physical property of all visible light determined by the light's frequency and visible to the human eye.
motikmotik
Color <span>is a physical property of all visible light determined by the light's frequency and visible to the human eye.</span>
6 0
2 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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2 years ago
An electrical conductor is an element with __________ electrons in its outer orbit.
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An electric conductor is an element with free electrons in its outer orbit
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2 years ago
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