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3 years ago

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According to Newtons second law of motion, which is equal to the acceleration of an object?A: net force + massB: net force x mas

sC: net force - massD: net force divides mass

2 answers:

Assoli18 [71]3 years ago

5 0

D: net force divides mass

Application of F = ma.

mart [117]3 years ago

4 0

Answer:

The acceleration of an object is equal to the net force divided by the mass of the object.

(D) is correct option.

Explanation:

According to Newton's second law of motion

The net force is equal to the product of the mass of the object and the acceleration of the object.

In mathematically terms,

$F = ma$

The acceleration is

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

Where, a = acceleration

F = net force

m = mass

Hence, The acceleration of an object is equal to the net force divided by the mass of the object.

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What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

kramer

I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

<span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

<span>x (max) = a(max) /ω² </span>

<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>

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3 years ago

A 0.500 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 m

KatRina [158]

To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.

Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.

Part A) Conservation of the energy

$mgh = \frac{1}{2} mv^2$

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(1.20)}$

$v_1 = 4.84m/s$

$\therefore v_1 = -4.84/s \rightarrow \text{Negative direction downward}$

Part B) Kinematic equation of linear motion,

$v_2^2 = u_0^2 +2a\Delta y$

Here

v= 0 Because at 1.5m reaches highest point, so v=0

$0 = u_2^2 +2(-9.8)(0.7)$

$u_2 = 3.7m/s$

Therefore the velocity after the collision with the floor is 3.7m/s

PART C) Total change of impulse is given as,

$J = P_2 -P_1$

$J = mU_2-mV_1$

$J = m(U_2-V_1)$

$J = (0.5)(3.7-(-4.84))$

$J = 4.27kg \cdot m/s \rightarrow \text{Upward because is positive}$

6 0

4 years ago

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

3 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

inysia [295]

The answer is B. 2.45 N/m.

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3 years ago

Which circle has minimum flux in the following figure?

Feliz [49]

Answer:

Circle Q

Explanation:

The flux of a field through a surface is given by:

$\Phi = F A cos \theta$

where

F is the field strength

A is the area of the surface

$\theta$ is the angle between the direction of the field and the normal to the surface

In the figure, we see that the 4 surfaces have different orientations, so different angle $\theta$ . In the formula, we see that the flux is maximum when $\theta=0^{\circ}$ , which occurs when the field is perpendicular to the surface, and the flux is zero when $\theta=90^{\circ}$ , (because $cos 90^{\circ}=0$ ) which occurs when the field is parallel to the surface.

In the figure, we see that circle Q is parallel to the direction of the field, so in this case the field is zero, while in all the other cases is not.

So, the correct answer is

Circle Q

6 0

4 years ago