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Vlad1618 [11]
2 years ago
15

The angle of deflection shows the amount of electric force on each balloon. A greater force causes the balloon to deflect to a g

reater angle. Since the two balloons are deflected to the same angle, we know that the force on each balloon must be the same. Why do you think the balloons have the same amount of force exerted on them?
Physics
1 answer:
UNO [17]2 years ago
7 0

Answer:

Hello the options to your question is missing attached below are the missing options

  • Electrical forces are symmetrical, The electric repulsion of the Balloons is equal because electric forces between two objects are always symmetrical
  • The electric forces are equal in this case because both balloons had the same amount of electric charge

Answer :

Electrical forces are symmetrical, The electric repulsion of the Balloons is equal because electric forces between two objects are always symmetrical

Explanation:

The Balloons have the same amount of force exerted on them, because Electrical forces are symmetrical, The electric repulsion of the Balloons is equal because electric forces between two objects are always symmetrical, therefore the two Balloons are deflected to the same angle

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a
Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

y = 0.049 \cos(7t) .......(1)

The general equation of a wave is given by :

y=A\cos(\omega t) .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0
2 years ago
A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 0
2 years ago
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0
2 years ago
un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente
Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

\Delta V = V_o\gamma \Delta T

here we know that

V_o = 40 Ltr

volume expansion coefficient of the gasoline is given as

\gamma = 950 × 10^{–6}

change in temperature is given as

\Delta T = (131 - 68) \times \frac{5}{9}

\Delta T = 35 ^oC

Now we have

\Delta V = 40(950 \times 10^{-6})(35)

\Delta V = 1.33 Ltr

3 0
2 years ago
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