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Acceleration = (change in speed) / (time for the change)

Change in speed = (New speed) - (old speed)

Change in speed = (0.8 m/s - 0.5 m/s) = 0.3 m/s

Time for the change = 3 seconds

Acceleration = (0.3 m/s) / (3 s)

Acceleration = 0.1 m/s²

(a hair more than 1 percent of 1 G.)

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

[tex] |F| = \frac{Kq_{1}q_{2}}{d^{2}} [/tex]

Where:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C  

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

[tex]|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N[/tex]                                      

Now, we can find the acceleration:

[tex] a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2} [/tex]

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!    

Analyzing the question:                                                                                        

We are given:

initial velocity (u) = 100 m/s

final velocity (v) = v m/s

distance (s) = 125 m

acceleration (a) = 5 m/s²

Solving for Final Velocity (v):                                                                              

from the third equation of motion:

v² - u² = 2as

v² - (100)² = 2(5)(125)

v² - 10000 = 1250

v² = 1250 + 10000

v² = 11250

v = 106.06 m/s

Answer:

Thank you ☺️

Answer:

The energy is   [tex]U =  18.98 \  J [/tex]

Explanation:

From the question we are told that

   The inductor is  [tex]L  =  4.74 \ H[/tex]

    The resistance of the resistor is [tex]R =  9.33 \  \Omega[/tex]

    The voltage of the battery is [tex]V =  26.4 \  V[/tex]

Generally the current flowing in the circuit is mathematically represented as

      [tex]I =  \frac{V}{R}[/tex]

=>   [tex]I =  \frac{26.4}{9.33 }[/tex]

=>   [tex]I =  2.83 \ A[/tex]

Generally the corresponding energy stored in the circuit is  

       [tex]U =  \frac{1}{2} * L  *  I^2[/tex]

        [tex]U =  \frac{1}{2} *  4.74  *  2.83 ^2[/tex]

       [tex]U =  18.98 \  J [/tex]

Answer:

700 N

Explanation:

The gravitational potential energy of the system decreases by 28,000 joules

The diver jumps from a height of 40 meters

Therefore the weight in Newton can be calculated as follows

= 28,000/40

= 700 newtons

Answer:

Answer is C or B

Explanation:

Answer:

C. Some social patterns are helpful, while others are harmful.

Explanation:

Hope this was helpful, Have an amazing,spooky Halloween!!

Answer:

578.66 N

Explanation:

The first step is to calculate the mass

mgh= 3200J

3200/9.8×5.53

3200/54.194

m = 59.047 kg

Therefore the weight can be calculated as follows

Weight = m × g

= 59.047 × 9.8

= 578.66 N

Answer:

a) and c)

Explanation:

        v = vₓ* x + vy* y

Answer:

The value is [tex]n=  9.375 *10^{15} \  electrons [/tex]

Explanation:

From the question we are told that

  The average current is  [tex]I  =  25.0 \mu A = 25.0 *10^{-6} \  A[/tex]

Generally the quantity of charge (electron )  is mathematically represented as

       [tex]Q =  ne[/tex]

Here e is the charge on a single electron with value [tex]e = 1.60  *10^{-19} \  C[/tex]

   Generally current is mathematically represented as

     [tex]I  = \frac{Q}{t}[/tex]

=>   [tex]I  = \frac{ne}{t}[/tex]

Here t is time which is given as 1 minutes =  60  seconds

  and  n is the number of electrons

So

      [tex]25.0 *10^{-6}  = \frac{ n* 1.60  *10^{-19}}{60}[/tex]

=>    [tex] 60  * 25.0 *10^{-6} =  n* 1.60  *10^{-19}  [/tex]

=>    [tex]n=  \frac{60  * 25.0 *10^{-6} }{ 1.60  *10^{-19} }[/tex]

=>    [tex]n=  9.375 *10^{15} \  electrons [/tex]

The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons

We are given;

Average Current; I = 25 μA = 25 × 10⁻⁶ A

Formula for the current is;

I = Q/t = ne/t

where;

n is number of electrons

e is electron charge = 1.6 * 10⁻¹⁹ C

t is time = 1 minute = 60 seconds

Thus making n the subject gives;

n = It/e

n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)

n = 9.375 × 10¹⁵ electrons

Read more about number of electrons at; https://brainly.com/question/11406294

Answer:

the current through 30 ohms resistor is 30 A.

Explanation:

Given;

resistor, R₁ = 10 ohms

resistor, R₂ = 30 ohms

voltage of the source = 120 V

The effective resistance, Rt = 30 ohms + 10 ohms = 40 ohms

The current through 30 ohms resistor is the same as the current through 10 ohms resistor because they are connected in series.

Appy ohms law to determine the currecnt through 30 ohms resistor;

I = V / Rt

I = 120 / 40

I = 3 A

Therefore, the current through 30 ohms resistor is 30 A.

Answer:

Explanation:

a) Using the equation of motion

v² = u²+2as

v is the final velocity = 0m/s

u is the initial velocity = 5m/s

S is the distance = 250m

a is the acceleration

Substitute given values:

0² = 5²+2a(250)

-25 = 500a

a = -25/500

a = -0.05m/s²

b) If a = 0.34m/s²

We need to get the initial speed u

v² = u²+2as

0 = u²+2(0.34)(250)

-u² = 170

u² = -170

u = √-170

The square root of a negative number will give a complex number, hence the speed is not possible

Answer:

D

Explanation:

Answer:

c.4s

Explanation:

Answer:

The minimum current in the rod that can cause it to float 2.96 A.

Explanation:

Given;

length of the copper rod, L = 1.43 m

mass of the rod, m = 236 g = 0.236 kg

magnetic field, B = 0.546 T

The magnetic of the rod is given by;

Fm = BIL

The downward force on the rod is given by;

F = mg

For the rod to float, the difference in the two force will be zero;

BIL - mg = 0

BIL = mg

I = mg / BL

I = (0.236 x 9.8) / (0.546 x 1.43)

I = 2.96 A

Therefore, the minimum current in the rod that can cause it to float 2.96 A.

Answer:

it's A trust me ok

Explanation:

i can't explain

Answer:

Explanation:

Accuracy can be said to mean the degree to which the particular result of a measurement, or calculation, and even possibly specification agrees or is the same with respect to the correct value or an established standard. Succinctly put, it's is how close a value is to the actual value it ought to be.

Precision on the other hand, is a change in a measurement, or calculation, and even as far as specification, much especially as represented by the number of digits that has been established. In other words, it is the proximity of two or more measurements with respect to one another.

Reproducibility occurs when a measurement(for example) is made by another person, or a different instrument is used. Yet, the same values are obtained.

They are very important in design because they account for very important part of an experiment. Neglecting these quantities means exposing an instrument to unknown danger to the factory and even the personnels.

Also, neglect in taking note of accuracy, precision and reproducibility can lead to poor data processing and even human errors.

Now, vote brainliest, will you? :)

Answer:

The time is [tex]t_t =   3.7583 \ s [/tex]

Explanation:

From the question we are told that

   The angle is  [tex]\theta =  30^o[/tex]

    The horizontal  distance is  [tex]d =  120 \ ft[/tex]

Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as

      [tex]v =  u  +  at[/tex]

here a =  -g  the negative sign is because the direction of motion is against gravity

So

       [tex]v =  v_i   +  at[/tex]

Here[tex] v_i [/tex] is the vertical  component of the initial  velocity of the ball which is  mathematically

represented as

        [tex]v_i = usin(\theta )[/tex]

So      

=>     [tex]0  =  usin(\theta )  -9.8t[/tex]

Generally the total time taken to travel the 120 ft  is mathematically represented as

        [tex]t_t = \frac{120}{v_h}[/tex]

Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as

     [tex]v_h  =  u cos(\theta )[/tex]

So

       [tex]t_t = \frac{120}{ u cos(\theta )}[/tex]

Generally the time taken to reach the maximum height is  

      [tex]t = \frac{t_t}{2}[/tex]

=>    [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]

=>    [tex]t = \frac{60}{ u cos(\theta )} [/tex]

So

      [tex]0  =  usin(\theta )  -9.8*   [\frac{60}{ u cos(\theta )}][/tex]

        [tex]  usin(\theta )   = 9.8*   [\frac{60}{ u cos(\theta )}][/tex]

       [tex]  usin(\theta ) *  u cos(\theta)  =60*  9.8   [/tex]

        [tex]  u^2 sin(\theta ) cos(\theta)  =60*  9.8   [/tex]

       [tex]  u^2 sin(30 ) cos(30)  =60*  9.8   [/tex]

        [tex]  u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2}  =588.6  [/tex]

         [tex]  u^2 *\sqrt{3}  =2354.4 [/tex]

          [tex]  u^2 = 1359.31 [/tex]

        [tex]  u = 36.87 \ ft/s  [/tex]

Substituting this value into the equation for total time

     [tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]

     [tex]t_t =   3.7583 \ s [/tex]

Answer:

For equilibrium, the mass m1 must be about 2.29 kg

Explanation:

The forces acting on m1 are : the weight (m1 * g) and the tension (T1) of the string.

The forces acting on m2 are:

1) along the ramp: the component of m2's weight (m2 g * sin(37)) and the tension (T2) of the string. Which by the way, must be equal in magnitude to T1 since the string is inextensible.

2) perpendicular to the ramp; the component of m2's weight (m2 g * cos(37)) and the normal from the contact with the ramp. These two compensate each other.

We therefore want the net force on each block to be zero for the system to be in equilibrium. This means:

T1 = m1 g

T2 = T1 = m2 g sin(37)

Then we have that if m2 = 3.8 kg, then:

m1 g = m2 g sin(37)

cancelling "g" on both sides:

m1 = m2 * sin(37) = 3.8 * sin(37) = 2.28689 kg

which may be rounded to about 2.29 kg.

Answer:

i. The magnetic force of repulsion.

ii. The magnetic force of attraction.

Explanation:

A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.

i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.

ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.

Answer:

a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].

b) The reading after 20 minutes is approximately 70.770 ºF.

Explanation:

a) Newton's law of cooling is represented by the following ordinary differential equation:

[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)

Where:

[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.

[tex]\tau[/tex] - Time constant, measured in minutes.

[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.

[tex]M[/tex] - Medium temperature, measured in Fahrenheit.

Now we proceed to solve the differential equation:

[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]

[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]

[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]

[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]

[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)

Where:

[tex]t[/tex] -Time, measured in minutes.

[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.

b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])

[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]

[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]

[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]

[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]

[tex]\tau = 5.461\,min[/tex]

The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:

[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]

[tex]T(20) \approx 70.770\,^{\circ}F[/tex]

The reading after 20 minutes is approximately 70.770 ºF.

Answer:

Options B & E are correct

Explanation:

Looking at all the options, B & E are the correct ones.

Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.

Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.

Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.

Option C is not correct because although the Electric field along the wire is not zero, it is very small.

Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :

b. Very little energy is dissipated in the thick connecting wires.

e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.

The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.

Thus, the correct answer is B and E.

Learn more about "Circuit":

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Answer:

250000 [m²]

Explanation:

A unit analysis has to be performed, where the unit of length is the meter. And the unit for the area is the square meter (m²)

L = 500 [m]

Therefore if we want to convert this length to meters we must square the length.

A = 500² = 500*500 = 250000 [m²]

Answer:

2I/7

Explanation:

Formula for moment of inertia through the centre of mass of a solid sphere is given as; I_sc = (2/5) mR²

Now, from parallel axis theorem, moment of inertia of solid sphere from tangent is given as;

I = (2/5) m R² + m R²

I = (7/5) mR²

Thus,mR² = 5I/7

Putting 5I/7 for mR² in first equation, we have;

I_sc = (2/5) × 5I/7

I_sc = 2I/7

Answer:

Arrector Pili Muscle - This is a tiny muscle that attaches to the base of a hair follicle at one end and to dermal tissue on the other end. In order to generate heat when the body is cold, the arrector pili muscles contract all at once, causing the hair to "stand up straight" on the skin.

Explanation:

Answer: be found in group 17 and be highly reactive

Explanation:

Elements are distributed in groups and periods in a periodic table.

Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.

Flourine, chlorine, bromine and iodine are elements which belong to Group 17. All of them contain 7 valence electrons each and need one electron to complete their octet.

The chemical reactivity of elements is governed by the valence electrons present in the element and thus all of them are highly reactive.

Complete Question

A diagram representing this question is shown on the first uploaded image

Answer:

The value is [tex]P = 294594.3 \ W[/tex]

Explanation:

From the question we are told that

   The height is  [tex]h  =  41 \  m[/tex]

   The efficiency of the turbine is [tex]\eta =  0.77[/tex]

   The flow rate is [tex]\r  V  = 1 m^3 / s[/tex]

    The head loss is  g =  2 m

Generally the head gain by the turbine is mathematically represented as      

        [tex]H  =  h -  d[/tex]

=>     [tex]H  =  41 - 2[/tex]

=>     [tex]H  =  39 \  m [/tex]

Generally the actual power generated by the turbine is mathematically represented as

          [tex]P =  \eta  *\gamma *    \r V * H[/tex]

Here [tex]\gamma[/tex] is the specific density of water with value

        [tex]\gamma   =  9810 N/m^3 [/tex]

So

       [tex]P =0.77  *9810 *   1 *  39[/tex]

       [tex]P = 294594.3 \ W[/tex]