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2 years ago

6

A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

de units in the answer to a numerical question.)

1 answer:

inysia [295]2 years ago

4 0

For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
Rewriting the acceleration as the increment of velocity $\Delta v$ in a time $\Delta t$ : $a= \frac{\Delta v}{\Delta t}$ , F becomes
$F=m \frac{\Delta v}{\Delta t}$
But given the definition of momentum: $p=mv$ , then $m \Delta v$ represents the momentum change. So we can rewrite F as
$F= \frac{\Delta p}{\Delta t}$
And re-arranging the formula we can calculate the value of the change in momentum:
$\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s$

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How do air mass conditions ahead of the squall line support the development of new cell?

IRISSAK [1]

<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>

7 0

2 years ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

3 0

2 years ago

In Burglar alarm LDR acts as a/an<br> a. off switch<br> b. on switch<br> c. AND gate<br> d. OR gate

Gelneren [198K]

In Burglar alarm, LDR acts an AND gate.

Answer: C

Explanation

The LDR is light dependent resistor. The principle used in the working of LDR is that the resistance is inversely proportional to the intensity of light falling on the diode.

In burglar alarm, LDR diode is combined with an IC 555.

Normally an LED source is made to be incident on the LDR diode with same intensity such that the resistance will be maintained constant.

As the LDR is connected with IC, the voltage will be high when light is falling on the diode.

The IC will give only two output states that is high and low. This confirms that LDR in burglar alarm act as AND gate.

As the thief enters and crosses the LED light, the intensity of the light falling on the diode will decrease leading to decrease in the voltage which will cause the alarm to beep.

4 0

2 years ago

A solid steel bar of circular cross section has diameter d 5 2.5 in., L 5 60 in., and shear modulus of elasticity G 5 11.5 3 106

8090 [49]

Answer:

A) θ = 4.9 x 10^(-3) rad

B) τ_max = 1.173 ksi

C) τ_a = 4.786 ksi

Explanation:

We are given;

diameter; d = 2.5 inches = 0.2083 ft

Length; L = 60 inches = 5 ft

Torque; T = 300 lb.ft

Shear modulus; G = 11.5 x 10^(6) psi = 11.5 x 144 x 10^(6) lb/ft² = 1.656 x 10^(9) lb/ft²

A) Now, formula to determine angle of twist is given as;

T/I_p = Gθ/L

Where I_p is polar moment of inertia

θ is angle of twist.

Now I_p = πd⁴/32 = π(0.2083)⁴/32 = 1.85 x 10^(-4) ft⁴

Thus, making θ the subject, we have;

TL/GI_p = θ

θ = (300 x 5)/(1.656 x 10^(9) x 1.85 x 10^(-4))

θ = 4.9 x 10^(-3) rad

B) Maximum shear stress is given by the formula ;

τ_max = (Gθ/L)(d/2)

From earlier, (Gθ/L) = T/I_p

Thus, (Gθ/L) = 300/1.85 x 10^(-4) = 1621621.6216

Thus,

τ_max = 1621621.6216 x (0.2083/2)

τ_max = 168891.89 lbf/ft²

Converting to ksi = 168891.89/144000 ksi = 1.173 ksi

C) Shear stress at radial distance is given as;

τ_a = (Gθ/L)•r_a

r_a is given as 5.1 inches = 0.425m

τ_a = 1621621.6216 x 0.425 = 689189.189 lbf/ft²

Converting to ksi = 689189.189/144000 ksi = 4.786 ksi

7 0

2 years ago

You have found a treasure map that directs you to start at a hollow tree, walk 300 meters directly north, turn and walk 500 mete

Dmitry_Shevchenko [17]

Answer:633 m

Explanation:

First we have moved 300 m in North

let say it as point a and its vector is $300\hat{j}$

after that we have moved 500 m northeast

let say it as point b

therefore position of b with respect to a is

r $_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}$

Therefore position of b w.r.t to origin is

$r_b=r_a+r_{ba}$

$r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}$

$r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}$

after this we moved 400 m $60^{\circ}$ south of east i.e. $60^{\circ}$ below from positive x axis

let say it as c

$r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}$

$r_c=r_{b}+r_{cb}$

$r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}$

$r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}$

magnitude is $\sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}$

=633.052

for direction $tan\theta =\frac{250\sqrt{2}+300-200\sqrt{3}}{250\sqrt{2}+200}$

$tan\theta =\frac{307.139}{553.553}$

$\theta =29.021^{\circ}$ with x -axis

7 0

2 years ago