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1 year ago

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a 0.13 kg stopper is attched to a 0.93 m length of string. the stopper is swung in.a horizontal circle making one revolution in

1.18 s find the speed of the mass find its centripetal accleration and find the force the string exerts on it

1 answer:

Crazy boy [7]1 year ago

8 0

The speed of the stopper is 4.95 m/s, while the centripetal force on the stopper and the tension in the string is 3.42N each.

The mass of the stopper is 0.13 Kg, the length of the string to which it is attached is 0.93 m. The stopper swung and made one revolution in 1.18 seconds, so the time period of revolution of the stopper is 1.18 seconds.

The relation between the time period and the angular speed is,

T = 2π/W

W is angular speed and T is the time period of revolution.

Putting values,

1.18 = 2π/W

W = 5.32 rad/s.

The speed of the stopper can be found as,

v = rW

r is the length of the string,

V = 4.95 m/s.

The centripetal force on the stopper is given by,

F = MV²/r

Putting values,

F = 0.13 x 4.95 x 4.95/0.93

F = 3.42 N.

The tension in the string will be equal to the centripetal force on the stopper, so, the tension is 3.

42 N.

To know more about centripetal force, visit,

brainly.com/question/20905151

#SPJ4

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3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration

Alexus [3.1K]

Answer:

a) a = 7.72 m / s², N = 19.9 N and b) F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

sin52 = Wx / W

cos52 = Wy / W

Wx = W sin52

Wy = w cos 52

Let's write them equations

X axis

Wx = ma

Y Axis

N-Wy = 0

N = Wy

a) Let's calculate the acceleration

a = W sin52 / m = mg sin 52 / m

a = g sin 52

a = 9.8 sin52

a = 7.72 m / s²

The force of the ramp is normal

N = Wy = mg cos 52

N = 3.3 9.8 cos 52

N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

F - Wx = 0

F = Wx

F = mg sin52

F = 3.3 9.8 sin 52

F = 25.5 N

Parallel to the plane and going up

3 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the win

nydimaria [60]

Answer:

Velocity

Explanation:

We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.

Hence, the given information describes wind velocity.

6 0

3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the

inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

Δx = 0.84 m

5 0

3 years ago