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3 years ago

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A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o

bject of mass m2 produces an acceleration of 1.80 m/s2.(a) What is the value of the ratio m1/m2?(b) If m1 and m2 are combined into one object, find its acceleration under the action of the force F with arrow.

1 answer:

Bas_tet [7]3 years ago

5 0

Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b) the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases, M₁a₁ = M₂a₂

$\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581$

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

$a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}$

$But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2$

Therefore, the acceleration of the combined masses is 1.139 m/s²

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A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distanc

geniusboy [140]

Answer:

a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b) The maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

a)

The total distance traveled by the sphere during one cycle of simple harmonic motion is:

d = 4A (where A is the Amplitude)

d = 4(3.40 cm)

d = 13.60 cm

Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b)

As we all know that:

$x = Asin \omega t$

Differentiating the above expression with respect to x ; we have :

$\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)$

$v = A \omega cos \omega t$

Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;

Then:

$v_{max} = A \omega$

We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium

substituting $2 \pi f$ for $\omega$ in the above expression;

$v_{max} = A(2 \pi f)$

$v_{max} = 3.40 cm (2 \pi *4.80)$

$v_{max} = 102.54 \ cm/s$

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

$v = A \omega cos \omega t$

By differentiation with respect to t;

$\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)$

$a =- A \omega^2 sin \omega t$

The maximum acceleration of the sphere is;

$a_{max} =A \omega^2$

where;

$w = 2 \pi f$

$a_{max} = A(2 \pi f)^2$

where A= 3.40 cm = 0.034 m

$a_{max} = 0.034*(2 \pi *4.80)^2$

$a_{max} = 30.93 \ m/s^2$

The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of $x = \pm A$

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As seen from the Earth, the Sun, Moon, and planets all appear to move along the ecliptic. ... Unlike the Sun, however, the planets don't always move in the same direction along the ecliptic. They usually move in the same direction as the Sun, but from time to time they seem to slow down, stop, and reverse direction!

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1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

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(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$

To learn more about projectile calculations please visit: brainly.com/question/14083704

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slava [35]

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2 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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