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3 years ago

8

6. Willingness to take turns is one way we can express our attitudes in

1 answer:

NNADVOKAT [17]3 years ago

5 0

The answer is A ..........

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When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking

postnew [5]

Answer:

$d=2.4\ miles$

Explanation:

Given:

average walking speed, $v_w=3\ mph$

average biking speed, $v_b=12\ mph$

<u>According to given condition:</u>

$t_w=t_b+\frac{36}{60}$

where:

$t_w=$ time taken to reach the building by walking

$t_b=$ time taken to reach the building by biking

We know that,

$\rm time=\frac{distance}{speed}$

so,

$\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}$

$\frac{d}{3}=\frac{d}{12} +\frac{3}{5}$

$d=2.4\ miles$

7 0

3 years ago

Read 2 more answers

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

b.

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

What is the difference between current and voltage besides their units of measurement

kondor19780726 [428]

"Voltage" is the "pressure" that makes electrons want to leave where they are
and head in some direction, if there's conducting material in that direction.

"Current" is the rate at which they all migrate in that direction.

8 0

3 years ago

Magnetic pole reversal worksheet

Likurg_2 [28]

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8 0

3 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago