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miss Akunina [59]

2 years ago

15

The mass of 04kg is rotated by a spring at a constant speed V . If the length of the string is 1m and the minimum tension is 3N,

what is value of V

1 answer:

wlad13 [49]2 years ago

3 0

I think you mean "rotated by a string", which is to say a 0.4-kg mass is suspended from a string and swung around in a circle with constant speed.

The centripetal acceleration a of the object is

a = V² / R

where R is the length of the string, 1 m, so that

a = V² / (1 m) = V² m⁻¹

As the mass goes through its uniform circular motion, the only relevant force acting on it is tension with magnitude T, pointing in the direction in which the mass is accelerating (the center of its circular path). By Newton's second law,

T = ma

where m is the object's mass, 0.4 kg. Then

3 N = (0.4 kg) (V² m⁻¹)

⇒ V² = (3 N) / (0.4 kg) • (1 m) = 7.5 m²/s²

⇒ V ≈ 2.74 m/s

You might be interested in

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

If an object moves twice as fast how do you think its energy changes? halves

il63 [147K]

Answer:

If an object moves twice as fast its kinetic energy quadruples.

Explanation:

The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:

K₁= 1/2 * m* v²

If we double the speed of the body, its kinetic energy (K₂) will be:

K₂= 1/2 * m*( 2v)²

K₂= 1/2 * m* 4 *v²

K₂= 4(1/2 * m *v²)

K₂= 4*K₁

6 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and

Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0

3 years ago

A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

3 years ago

A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

Zina [86]

Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0

3 years ago