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3 years ago

15

Choose all the answers that apply. Which of the following forces require physical contact? tension, air resistance, friction, ma

gnetic force, applied force

1 answer:

svet-max [94.6K]3 years ago

8 0

Answer:

I belive it would be friction applied force and tension

Explanation:

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A cross-country runner accelerates from 4 m/s to 7 m/s over the course of 5 seconds. What is the runner's average rate of accele

fredd [130]

B. 0.6 for show
hopefully this works lemme know

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3 years ago

How far will a bus travel if it averages a speed of 65 km/h for 7 hours?

otez555 [7]

Answer:

the bus travles 65x7=455 km

Explanation:

hope this helps you

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2 years ago

Read 2 more answers

A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving a

emmainna [20.7K]

Answer:

a)

0.245 m/s

b)

0.904 m

Explanation:

a)

$v_{d}$ = speed of duck ahead of wave

$v_{s}$ = speed of surface wave = 0.32 m/s

T = time for paddling = 1.6 s

d = spacing between the waves = 0.12 m

speed of duck ahead of wave is given as

$v_{d}$ = $v_{s}$ - $\frac{d}{T}$

$v_{d}$ = 0.32 - $\frac{0.12}{1.6}$

$v_{d}$ = 0.245 m/s

b)

$v_{w}$ = speed of wave behind the duck

speed of wave behind the duck is given as

$v_{w}$ = $v_{s}$ + $v_{d}$

$v_{w}$ = 0.32 + 0.245

$v_{w}$ = 0.565 m/s

D = spacing between the crests

spacing between the crests is given as

D = $v_{w}$ T

D = (0.565) (1.6)

D = 0.904 m

7 0

3 years ago

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a

xxTIMURxx [149]

Answers:

a) $F_{g}=735 N$ and $n=732.47 N$ , hence $F_{g} > n$

b) $n_{poles}=735 N$ $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the <u>centripetal force</u> $F_{c}$ and the <u>gravitational force</u> $F_{g}$ (also called true weight) are directed "downward", while the <u>normal force</u> $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation:

$n_{equator}-F_{g}=-F_{c}$ (1)

Where:

$F_{g}=m g$ being $m=75 kg$ the mass and $g=9.8 m/s^{2}$ the acceleration due gravity

$F_{c}=m a_{c}$ being $a_{c}=0.0337 m/s^{2}$ the centripetal acceleration at the equator

According to this (1) is rewritten as:

$n_{equator}-mg=-m a_{c}$ (2)

Isolating $n_{equator}$ :

$n_{equator}=-m a_{c} + mg$ (3)

$n_{equator}=m(-a_{c}+g)$ (4)

$n_{equator}=75 kg (-0.0337 m/s^{2}+9.8 m/s^{2})$ (5)

$n_{equator}=732.47 N$ (6) This is the apparent weight at the equator

The true weight is given by $F_{g}=m g=75 kg (9.8 m/s^{2})$

Hence: $F_{g}=735 N$ (7)

As we can see $F_{g} > n_{equator}$

b) Now we have to calculate the apparent weight at the poles $n_{poles}$ :

$n_{poles}-F_{g}=-F_{c-poles}$ (8)

Since $F_{c-poles}=0$ (8) is rewritten as:

$n_{poles}=F_{g}$ (9)

$n_{poles}=m g$ (10)

$n_{poles}=(75 kg)(9.8 m/s^{2})=735 N$ (11)

So, the apparent weight of the person at the poles is 735 N and at the equator is 732.47 N

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3 years ago

How long would it take someone jogging at 2.4 m/s to travel 2300 m?

Ber [7]

Answer:

958 Second or 15 minute and 58 second approximately 16 minutes

Explanation:

m / m/s

2300/2.4 = 958 second

958/60 = 15 minutes and 58 seconds approximately 16 minutes

7 0

2 years ago