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AURORKA [14]
1 year ago
10

Carts, bricks, and bands

Physics
1 answer:
tatiyna1 year ago
6 0

The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.

<h3>What is acceleration?</h3>

Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).

From the table given,

Trial 1 ----> 1 band = 0.24m/s²

Trial 2 ----> 2 bands = 0.51 m/s²

Trial 3 ----> 3 bands = 0.73 m/s²

Trial 4 -----> 4 bands = 1.00 m/s²

This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.

This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.

Learn more about acceleration here:

brainly.com/question/25749514

#SPJ1

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

4 0
3 years ago
What is the wave speed of a 6 m wave with a frequency of 20 Hz?
andrew-mc [135]

Answer: Wave speed= frequency x wavelength

=20 x 3

=60 m/s

Explanation:

8 0
3 years ago
An electron is released form rest in a region of space where a uniform electric field is present. Joanna claims that its kinetic
djyliett [7]

Answer:

Neither.

Explanation:

When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).

It will move towards a point  with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:

⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)

As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.

4 0
3 years ago
Alguien me ayuda con calculos estequiometricos?
ruslelena [56]

Answer:

SURE!!!...

But what to calculate!!!....

5 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
2 years ago
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