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ololo11 [35]
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?

Physics
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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There are nine PCNs in the Edmonton area. They work alongside more than 1,100 family doctors in over 330 clinics to provide care for 1.2 million patients. PCN teams include more than 370 nurses, mental health clinicians and other health professionals.

Explanation:

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All chemical reactions occur at the same rate true or false​
SSSSS [86.1K]

Answer:

false

Explanation:

7 0
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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

#SPJ1

3 0
11 months ago
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Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

6 0
2 years ago
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