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3 years ago

Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force

1 answer:

7 0

I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.

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How much support force does a table exert on a book that weighs 15 N when the book is placed on the table?

mestny [16]

Answer:

15 N

Explanation:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. This reaction is equal in magnitude to the force acting but in an opposite direction.

Now, if the book weighs 15 N, an opposite equal force will be: N = -15 N

But the magnitude of this will be the absolute value which is 15N.

8 0

3 years ago

If F1 = 104 and F2 = 104, what will be the net force?

ladessa [460]

Answer:

208

Explanation:

add it together for the answer

3 0

3 years ago

A seismographic station receives S and P waves from an earthquake, separated in time by 17.8 s. Assume the waves have traveled o

Georgia [21]

Answer:

D = 230.2 Km

Explanation:

let distance between seismograph and focus of quake is D

From time distance formula we can calculate the time taken by the S wave

$T_1 =\frac{D}{4.5}$

From time distance formula we can calculate the time taken by the P wave

$T_2 =\frac{D}{6.90}$

It is given in equation both waves are seperated from each other by 17.8 sec

so we have

$T_1 - T_2 = 17.8$ sec

Putting both time value to get distance value

$\frac{D}{4.5} - \frac{D}{6.90} = 17.8$

D = 230.2 Km

8 0

3 years ago

A football team wears a yellow shirt and white shorts. What would the kit look like:

AysviL [449]

In all three cases . . .

-- the shirt appears black, since there's no yellow light shining on it;

-- the shorts appear to have the same color as the light shining on them.

4 0

3 years ago

An air tank of volume 1.5 m3 is initially at 800 kPa and 208C. At t 5 0, it begins exhausting through a converging nozzle to sea

Serggg [28]

Answer:

(a) m = 0.141 kg/s

(b) t = 47.343 s

Explanation:

Given that:

The volume of air in the tank V = 1.5 m³

The initial temperature in the tank is supposed to be 20° C and not 208 C;

So $T_o = 20^0 C = ( 20 +273) K = 293K$

The initial pressure in the tank $P_o= 800 \ kPa$

The throat area $A_t$ = 0.75 cm²

To find the initial mass flow in kg/s.

Lets first recall that:

Provided that $\dfrac{P_{amb}}{P_{tank}}< 0.528$ , then the flow is choked.

Then;

$\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}$

$\dfrac{P_{amb}}{P_{tank}}= 0.1266$

From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.

Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:

$m = \rho \times A \times V$

where;

$\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}$

$\rho =9.51350( 0.8333 ) ^{2.5}$

$\rho =6.03 \ kg/m^3$

$T = T_o \bigg ( \dfrac{2}{k+1}\bigg)$

where;

$T_o = 293 \ K$

$T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)$

$T = 293 \bigg ( \dfrac{2}{2.4}\bigg)$

$T = 293 ( 0.8333)$

T = 244.16 K

During a critical condition when Mach No. is equal to one;

$V = a = \sqrt{kRT}$

$V = \sqrt{1.4 \times 287 \times 244.16}$

$V = \sqrt{98103.488}$

V = 313.214 m/s

Thus, the initial mass flow rate $m = \rho \times A \times V$

m = 6.03 × 0.75 × 10⁻⁴ × 313.214

m = 0.141 kg/s

(b)

The mass balance formula for the control volume surrounding the tank can be expressed as:

$\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)$

$= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m$

When the air mass flow rate is:

$m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

Thus; replacing $m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$ in the previous equation; we have:

$\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

$\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt$

Taking the differential of both sides from 0 → t

$In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t$

$In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t$

$\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$

So, when the pressure P = 500 kPa, the time required is:

$\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

t = 47.343 s

(c)

Let us recall that:

The choking on the nozzle occurred when $\dfrac{P_{amb}}{P_{tank}} = 0.528$

$\dfrac{P_{amb}}{0.528} = P_{tank}$

$\dfrac{101.35}{0.528} = P_{tank}$

$P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa$

From $\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$ ; the time required for $P_{tank} \simeq 192 \ kPa$ is:

$\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

By solving:

t = 143.745 s

7 0

3 years ago