a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​

Answer:

Alkaline batteries have longer shelf life than other batteries of the chloride type electrolyte batteries. They have a much higher energy density compared with other batteries. This allows the battery to produce the same energy while lasting longer than other batteries.

Advantage

1) Alkaline battery lasts long than other batteries  

2) It have a  much higher energy density

Disadvantage :

1) They are heavier and bulkier

2) They have high internal resistance

An alkaline battery is a type of primary battery that derives its energy from the reaction between zinc metal and manganese dioxide.

Advantage

Alkaline battery   can produce same amount of energy by lasting longer than other batteries  as it have longer shelf life than  lead acid accumulator​

Alkaline battery  have a much higher energy density compared to lead acid accumulator

Disadvantage :

Alkaline battery  are heavier and bulkier

Alkaline battery   have high internal resistance

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Answer:

v = 333.26 m/s

Explanation:

Given that,

The frequency of the tuning fork, f = 262 Hz

It produces the third resonance position above a closed air column that is 1.59 m in length.

We need to find the speed of sound in tuning fork. Let it is f. The third resonance position corresponds to the fifth harmonic of the closed air column. The wavelength in third resonance is given by :

[tex]\lambda=\dfrac{4}{5}L\\\\=\dfrac{4}{5}\times 1.59\\\\=1.272\ m[/tex]

The speed of sound is :

[tex]v=f\lambda\\\\=262\times 1.272\\\\=333.26\ m/s[/tex]

So, the speed of sound is 333.26 m/s.

Answer:

The speed of sound is 555.44 m/s.

Explanation:

frequency of third resonance position, f''' = 262 Hz

length of pipe, L = 1.59 m (closed)

Let the speed is v.

The frequency of third resonance in closed organ pipe is

[tex]f''' =\frac{3 v}{4 L}\\\\262 = \frac{3 v}{4\times 1.59}\\\\v = 555.44 m/s[/tex]

Answer:

A. It increased gradually as the reaction progressed.

Answer: A it increased gradually

as the reaction progressed

Answer:

Too what i know its because of the air

Explanation:

The steel rod is more dense and doesnt have air whil'st the oil tanker is less dense with lots of air

Answer:

No, it's not there.

Explanation:

For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.

Answer:

C

Explanation:

Answer:

6.34×10⁹ seconds

Explanation:

Applying,

V = d/t............. Equation 1

Where V = speed plate, d =  distance covered by the plate, t = time it takes the plate to travel

make t the subject of the equation

t = d/V............ Equation 2

From the question,

Given: d = 5 cm/yr = (5×3.154×10⁻⁹) = 1.577×10⁻⁸ m/s,  d = 100 meter

Substitute these values into equation 2

t = 100/1.577×10⁻⁸

t = 6.34×10⁹ seconds

Hence the time it takes the the plate to travel is 6.34×10⁹ seconds

Answer:

11760 J

Explanation:

cuz potential engery is PE = MHG

SO 80×15×9.8= 11760 J

Answer:

11769j

Explanation:

here,

mass(m)=80 kg

height(h)=15m

acceleration due to gravity(g)=9.8m/s^2

now,

potential energy = m×g×h

= 80×9.8×15

= 11760j

Explanation:

here's the answer to your question

Answer:

The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.

Explanation:

Let suppose that both lion and the Thomson's gazelle collide each other inelastically, the use of the Principle of Linear Momentum Conservation suffices to describe the entire phenomenon:

[tex]m_{L}\cdot \vec v_{L} + m_{T}\cdot \vec v_{T} = (m_{L}+m_{T})\cdot \vec v[/tex] (1)

Where:

[tex]m_{L}[/tex] - Mass of the lion, in kilograms.

[tex]m_{T}[/tex] - Mass of the Thomson's gazelle, in kilograms.

[tex]\vec v_{L}[/tex] - Initial velocity of the lion, in meters per second.

[tex]\vec v_{T}[/tex] - Initial velocity of the Thomson's gazelle, in meters per second.

[tex]\vec v[/tex] - Final velocity of the lion-gazelle system, in meters per second.

Let suppose that both northward velocity and eastward velocity are positive.

If we know that [tex]m_{L} = 109\,kg[/tex], [tex]\vec v_{L} = (0, 22.222)\left[\frac{m}{s} \right][/tex], [tex]m_{T} = 36\,kg[/tex] and [tex]\vec v_{T} = (21.667, 0)\,\left[\frac{m}{s} \right][/tex], then the final velocity of the lion-gazelle system is:

[tex]109\cdot (0,22.222)+36\cdot (21.667,0) = 147\cdot (v_{x},v_{y})[/tex]

[tex](v_{x}, v_{y}) = (0, 16.478) + (5.306, 0) \,\left[\frac{m}{s} \right][/tex]

[tex](v_{x}, v_{y}) = (5.306, 16.478)\,\left[\frac{m}{s} \right][/tex]

[tex](v_{x}, v_{y}) = (19.102, 59.321)\,\left[\frac{km}{h} \right][/tex]

And the final speed of the lion-gazelle system is calculated by the Pythagorean Theorem:

[tex]v = \sqrt{19.102^{2}+59.321^{2}}\,\left[\frac{km}{h} \right][/tex]

[tex]v \approx 62.321\,\frac{km}{h}[/tex]

The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.

Answer:

A simple siphon raises water over a crest and discharges it at a lower level.It works coninueasly due to pull of gravity.

Explanation:

Siphons are tubes which draw fluid over the rim of a tank to a lower point. After an initial pressure change to initiate the flow, siphons operate continuously due to the pull of gravity.

Working:

A simple siphon raises water over a crest and discharges it at a lower level. As water flows through a siphon, energy due to pressure and elevation is either lost to pipe friction or converted to velocity energy.

Answer:

CG gh sure er go b vh pxuh FPI OO c AM h kh

Answer:

velocity

Explanation:

this is because velocity has both magnitude and direction

Answer:

Explanation:

There is no pressure of your feet on the scales, and no pressure of the floor on the scales, so the scales will read zero*. Hence, your weight, in a freely falling lift is zero

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Conversion:

100 cm = 1 m

X cm = 0.75 m

Cross-multiplying, we have;

X = 0.75 * 100 = 75 cm

First of all, we would find the effort arm;

Effort arm = length of crow bar - length of load arm

Effort arm = 200 - 75

Effort arm = 125 cm

Next, we would determine the mechanical advantage (M.A) of the crow bar;

[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]

Substituting the values into the formula, we have;

[tex] M.A = \frac {125}{200} [/tex]

M.A = 0.625

To find the effort of the crow bar;

[tex] M.A = \frac {Load}{Effort} [/tex]

Making "effort" the subject of formula, we have;

[tex] Effort = \frac {Load}{M.A} [/tex]

[tex] Effort = \frac {600}{0.625} [/tex]

Effort = 960 Newton

Lastly, we would determine the velocity ratio (V.R);

[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]

[tex] V.R = \frac {125}{75} [/tex]

V.R = 1.67

Assuming the applied force is exerted parallel to the floor, by Newton's second law both the net vertical and net horizontal forces would be zero:

∑ F (horizontal) = p - f = 0 … … … because the desk is pushed at a constant speed

∑ F (vertical) = n - mg = 0 … … … because the desk doesn't move up or down

where p is the magnitude of the applied force, f is the mag. of kinetic friction, n is the mag. of the normal force due to contact between the floor and desk, and mg is the weight of the desk. We have

n = mg = (100 kg) g = 980 N

and the mag. of friction is proportional to n according to

f = 0.4 n = 392 N

Then the applied force p has magnitude

p = f = 392 N

i believe it would be Star A since Point A is a dwarf star and dwarf stars are usually most likely to have red coloring

Answer:

Star A

Explanation:

Literally the question has provided the answer itself

Calculation not required.

See the wordings

First star is dwarf star .

Dwarf star is mostly reddish brown or red colored .

Hence star A is the answer

Answer:

(a)  333.77 J

(b) 237.85 J

(c) 4763.77 J

(d) 4667.85 J

Explanation:

Temperature of source, TH = 314 K

Temperature of A, Tc = 292 K

Temperature of B, Tc' = 298 K

heat taken out, Qc = 4430 J

Let the heat deposited outside is QH and QH' by A and B respectively.

[tex]\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J[/tex]

Now

[tex]\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J[/tex]

(a) Work done for A

W = QH - QC = 4763.77 - 4430 = 333.77 J

(b) Work done for B

W' = QH' - Qc = 4667.85 - 4430 = 237.85 J

(c) QH = 4763.77 J

(d) QH' = 4667.85 J

Answer:

[tex]X=3.37561*10^{-16}[/tex]

Explanation:

From the question we are told that:

Width [tex]w=2.7m[/tex]

Acceleration [tex]a=9.8m/s^2[/tex]

Generally the equation for The apparent deflection is mathematically given by

X= 0.5 *g*t^2

Where

[tex]t=\frac{width}{Velocity of light}[/tex]

[tex]t=\frac{2.5}{3*10^8}[/tex]

[tex]t=8.3*10^{-9}[/tex]

Therefore

[tex]X= 0.5 *9.8*(8.3*10^{-9})^2[/tex]

[tex]X=3.37561*10^{-16}[/tex]

Answer:

Explanation:

i) frequency is the square root of the quotient of Tension over linear density/ all divided by two times the length.

linear density and tension are constant, so the numerator remains constant

The length is doubled, so the frequency is halved.

f = 128 Hz

ii) Doubling the tension increases the frequency by √2

f = 256√2 = 362 Hz

2) If other factors remain constant, increasing tension increases frequency.

Answer:

w = 0.01 t⁻²   rad/h

Explanation:

Let's use trigonometry to find the elevation angle of the telescope, we assume that the rocket remains fixed in the telescope.

        tan θ = CO / CA

the adjacent leg is CA = 16 km = 16 10³ m

Let's use kinematics to find the height of the rocket

          v = y / t

          y = v t

this height is equal to the opposite leg

          CO = v t

we substitute

         tan θ = vt / 16

         tan θ = 1200/16 t

         tan θ = 75 t

         θ = tan⁻¹ ( 75 t )

speed is defined by

          w = dθ/dt

          w = [tex]\frac{1}{1 + (75 t)^2} \ 75[/tex]  

as time increases we can neglect the 1 of the denominator

          w = 1/75 t²

          w = 0.0133 t⁻²

Using three significant figures

           w = 0.01 t⁻²   rad/h

for this answer the time must be given in hours

Answer: i)A to B : (ice) freezing

ii) B to C (water) boiling

C to D  (steam) evaporating

explanation:  0° is the freezing point of water  when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water  completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

Answer:

334m/s

Explanation:

If you take the data you gave and insert it into the equation f=v/λ the frequency would come out to be 3.34m

334 m/s is  the Velocity of sound in air.

Wave velocity in common usage refers to speed, although, properly, velocity implies both speed and direction.

The velocity of a wave is equal to the product of its wavelength and frequency (number of vibrations per second) and is independent of its intensity.

Mathematically , v=fλ

According to the question,

Sound frequency (f)  = 100Hz

Wavelength (λ)= 3.34m

Computing the values in the formula,

v=fλ

v = 100Hz x 3.34m

v= 334m/s

Therefore,

Velocity of sound in air is 334 m/s.

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Answer:

Satellites don't fall from the sky because they are orbiting Earth. Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity--combined with the satellite's momentum from its launch into space--cause the satellite go into orbit above Earth, instead of falling back down to the ground.