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3 years ago

15

A net force of 25.0 N causes an object to accelerate at 4.00 m/s2. What is the mass of the object?

2 answers:

maxonik [38]3 years ago

3 0

Answer:

6.25kg

Explanation:

m=F/a = 25/4 = 6.25kg

defon3 years ago

3 0

Answer:

Mass, m = 6.25 kg

Explanation:

It is given that,

Net force acting on the object, F = 25 N

Acceleration of the object, $a=4\ m/s^2$

Let m is the mass of the object. Using Newton's second law of motion. The force is given by :

F = ma

$m=\dfrac{F}{a}$

$m=\dfrac{25\ N}{4\ m/s^2}$

m = 6.25 kg

So, the mass of the object is 6.25 kg. Hence, this is the required solution.

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Read 2 more answers

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

2 years ago

The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.

ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

v is speed of sound
L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f = v/(4L)

251.1 = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

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11 months ago

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Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

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2 years ago