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bija089 [108]
2 years ago
8

A delivery truck starts it’s run by driving 5.20 km due west before turning due north and driving an additional 2.10 km. Finally

, the truck turns 30.0 degrees north of east and drives for 3.70 km before reaching its first dropoff point. What is the magnitude of the total displacement of the truck from where it started to its first dropoff point?
Physics
1 answer:
Gnoma [55]2 years ago
8 0

Answer:

4.427 m

Explanation:

We shall consider east as + x- axes and north as + ve y- axes. .

We shall represent every displacement in vector form as follows

D₁ = 5.2 km due west = - 5.2 i

D₂ = 2.1 km due north = 2.1 j

D₃ = 3.7 km towards north east at 30 degree from east

= 3.7 cos30 i + 3.7 sin 30 j = 3.2 i + 1.85 j

Total displacement D = D₁ + D₂ +D₃ +D₄.

- 5.2 i + 2.1 j + 3.2 i + 1.85 j

= - 2 i + 3.95 j

Magnitude of D

D² = (2)² + (3.95)²

= 4 + 15.6025

D = 4.427 m

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To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

t = \frac{d}{v}

t = \frac{250km}{95km/h}

t = 2.63h

Time taken for the lunch is

t = 1h

The time taken travel a distance of 250km with a speed of 55km/h

t = \frac{d}{v}

t = \frac{250}{55}

t = 4.54h

The total time taken is

t = t_{outgoing}+t_{lunch}}t_{return}

t = 2.63+1+4.54

t = 8.17h

The average speed is the ratio of total distance and total time

v = \frac{250+250}{8.17}

v = 61.15km/h

PART B)

As the displacement is zero the average velocity is zero.

5 0
2 years ago
Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

P = 18 W

7 0
2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0
2 years ago
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Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an
gulaghasi [49]

Complete Question

The complete question is shown on the first uploaded image

Answer:

  The correct answer is option c

Explanation:

Faraday states that when there is a change in magnetic field of a coil of a wire, it means that there exist an emf in the circuit which in induced due to the change in the magnetic flux

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This magnetic field that is produce would cause a change flux which would induce current in the second coil so the ammeter would indicate current flow in the second coil

a is incorrect because the current in fir coil is not change hence flux won't change therefore current is is not induced in second coil

This is the same reason b is incorrect

d is incorrect due to the fact that when the second coil is connected to a power supply by rewiring it to be in series with first coil the law of electromagnetism would no longer hold so he ammeter would show no reading  

6 0
2 years ago
A spring stores 10. joules of elastic potential
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The answer would be . Since we are looking for the spring constant you would need to use the formula
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. Then you'd substitute, for PEs and x.
10j=1/2k(.20m^2).
Then solve. k=500n/m
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2 years ago
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