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Contact [7]
2 years ago
13

1. A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15mins, wh

at is the average speed of the bus (m/s)? 2. Using the information from the previous question, what is the average velocity of the city bus?
Physics
1 answer:
loris [4]2 years ago
4 0

Answer:

<em>1) 1.56 m/s</em>

<em>2) 1.11 m/s</em>

<em></em>

Explanation:

The bus travels 6 blocks east and 8 blocks west

Each block is 100 m long

time taken to travel through this distance = 15 min

average speed of the bus = ?

the total blocks traveled = 6 + 8 = 14 blocks

total distance traveled = 14 x 100 m = 1400 m

time taken = 15 min = 15 x 60 sec = 900 sec

average speed of the bus = distance traveled/time taken

==> 1400/900 = <em>1.56 m/s</em>

<em></em>

2) velocity = displacement/time taken

displacement is the shortest distance between the starting position and the final position of the bus.

The displacement of this bus will be the hypotenuse of the triangle formed by the motion of the bus

the distance traveled east = 6 x 100 = 600 m

the distance traveled north = 8 x 100 = 800 m

displacement = \sqrt{800^2 + 600^2} = 1000 m

The velocity of the bus = 1000/900 = <em>1.11 m/s</em>

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A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast
Neko [114]

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

8 0
2 years ago
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For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to
photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

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2 years ago
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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
2 years ago
A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans
sertanlavr [38]

Answer:

0.4 A

Explanation:

From the question,

Electric power = Voltage×current

P = VI.......................... Equation 1

Make I the subject of the equation

I = P/V..................... Equation 2

Given: P = 96 J/s, V = 230 V.

Substitute into equation 2

I = 96/230

I = 0.4 A.

Hence the current is 0.4 A

8 0
2 years ago
The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac
Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

Pressure = h\cdot\rho\cdot g

Case 1 - Mercury in the tube

Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1

Case 2 - Water in the tube

Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2

Since atmospheric pressure is same

.P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g

or,  h_2=\frac{h_1\rho_1}{\rho_2}

Given\ h_1= 0.76\  m,\rho_1=13.6\cdot\rho_2

∴ h_2=0.76\cdot13.6=10.336\ m

Hence height of the water column =10.336 m

6 0
2 years ago
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