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3 years ago

14

Grade 11 physical science experiments on newtons second law experience

1 answer:

Nutka1998 [239]3 years ago

8 0

Answer:

Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass . Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.

Explanation:

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How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al

dusya [7]

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

$Q = ms\Delta T$

here we know that

m = 50 g

$\Delta T = 110 - 10$

$\Delta T = 100 ^oC$

so we have

$Q = 50(0.95)(100)$

$Q = 4750 J$

5 0

3 years ago

A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Tomtit [17]

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

6 0

2 years ago

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The materials carried by the circulatory system include which of the following? (1) blood, (2) hormones, (3) oxygen, (4) cellula

Mariana [72]

Well it does carry blood and oxygen! Sorry but I'm not sure about 2 and 4.

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2 years ago

Read 2 more answers

In the following equation , which compounds are the reactants? NaCL+AgNo>NaNo+AgCl

NeX [460]

The first two are always the reactants the products come after so they are last

3 0

2 years ago

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

3 years ago