A ball of mass 50 g is suspended at a height of 2.5 m from the ground. In terms of the gravitational
potential energy (GPE) and kinetic energy (KE) of the ball, which of the following statements is
correct?

A GPE is maximum, KE is 0
B. KE is maximum, GPE is 0
C. GPE = KE at this point
D. the ball has no GPE
​

Answer:

So that it may trap the solar radiations and maintain the temperature of the wooden building.

Explanation:

Wood is not as effective an insulator as fiberglass. It comes in a number of widths, the widest being 3/4 inch, which has an R-value of 0.94. However, fiberglass, the thinnest of which is 3/4 inch, has an R-value of 3. Therefore, fiberglass is a better insulator.

Answer:

The beauty of science is everywhere.

Explanation:

science is extremely important and you see it all around you. For instance you need knowledge of chemistry to create medication in order to help cure the sick. Knowledge of physics is needed in order to construct building or vehicles.

make sure to ask if you need anymore examples or proof :)

Answer:

B. Charges that move.

The specific heat of the metal, assuming no heat is exchanged with the surroundings is 2140 J/(kg•K).

The specific heat capacity of the metal is determined from the principle of conservation of energy.

energy lost by the metal = energy gained by aluminum + energy gained by water

Q = mcΔθ

where;

0.425c(100 - 40) = 0.1(900)(40 - 15) + 0.5(4186)(40 - 15)

25.5c = 2250 + 52,325

c = 54,575/25.5

c = 2140 J/(kg•K)

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Answer:

elastic potential energy

You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.

Explanation:

The distance traveled by the pool ball is 0.376 m.

To calculate the distance traveled by the ball, we need to first find the resultant velocity of the ball.

This is the single velocity obtained when two or more velocities are combined.

To calculate the resultant velocity, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Finally, to calculate the total distance, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute into equation 2

Hence, the distance traveled by the pool ball is 0.376 m.

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The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C

Reading in kelvin scale = 252.5 K

The relationship between Fahrenheit and celsius can be expressed as

F = (9/5)(C) + 32 ---- ( 1 )

Therefore the temperature at which Fahrenheit scale quarters the celsius scale is :

F = 1/4 C --- ( 2 )

Back to equation ( 1 )

1/4 C = (9/5)(C) + 32

1/4C - 9/5C = 32

C ( 1/4 - 9/5 ) = 32

C = 32 / ( 1/4 - 9/5  )

   = -20.64⁰C

F = 1/4 ( -20.64 ) = -5.16 ⁰F

Hence we can conclude that The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C. Reading in kelvin scale = 252.5 K

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a) If you’re heading due North, total acceleration will be 0.0478 m/s².

b) If you’re heading due East, acceleration will be 0.0803 m/s².

c) If you’re heading straight up, no exact value for acceleration.

The time rate of change of velocity is known as acceleration.

a)If the boat is going along North from equator, there will be only centrifugal force acting along upward direction which will be balanced by the weight of boat.

Angular velocity ω= (2π / (3600 x 24)) = 7.27 x 10⁻⁵ rad/s

Acceleration due to centrifugal force a = 0.0337m/s²

Centrifugal force due to its own velocity 300 m/s is v²/r = 0.0141 m/s².

Total acceleration will be 0.0141 +0.0337 =0.0478m/s².

Thus, if you’re heading due North, total acceleration will be 0.0478 m/s².

b) If the boat goes along East, Coriolis acceleration is in the direction of centrifugal force.

Coriolis acceleration is (2 x 7.27 x 10⁻⁵ x300 ) = 0.0436m/s².

Total centrifugal acceleration due to centrifugal force = 300 +  (7.27 x 10⁻⁵ x 6371 x 10³) = 0.0436 m/s².

Total acceleration is 0.0436 + 0.0367 = 0.0803m/s².

Thus, if you’re heading due East, acceleration will be 0.0803 m/s².

c) When boat is going upward, centrifugal force will change with time. but Coriolis acceleration remains same.

Thus, if you’re heading straight up, no exact value for acceleration.

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Answer:

Hindi ko Po alm sorry Po talaga

The tension in the string is same in both cases because of the similar mass or weight of the bodies.

Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop, then there is same tension occurs in both cases because tension depends on mass which is similar on both side. We know that tension is the opposite force of the weight.

So we can conclude that the tension in the string is same in both cases because of the similar mass or weight of the bodies.

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Answer:

The pickle bottle cap on dipping it in hot water expanded.

Explanation: Hope it helps you:))))

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Answer:

Watts

Explanation:

A Watt is the unit of electrical power equal to one ampere under the pressure of one volt.

Answer:

Period becomes 1/3 of the original

Explanation:

Period = 1/f      if f becomes 3f   then period becomes 1/3 of the original

Answer:

Explanation:

X-rays are produced due to sudden deceleration of fast-moving electrons when they collide and interact with the target anode. In this process of deceleration, more than 99% of the electron energy is converted into heat and less than 1% of energy is converted into x-rays

The difference in the speeds of the two balls when they hit the ground is mathematically given as

V=0.10m/s

Generally, the equation for Acceleration due to  electric field is mathematically given as

[tex]A=g+\frac{q1E}{m}\\\\Therefore\\\\A=9.8*\frac{600*1000*150*}{0.520}\\\\A=9.97m/s2[/tex]

For Negative charge

[tex]A=g+\frac{q1E}{m}\\\\A=9.8*\frac{-600*1000*150*}{0.520}\\\\A=9.62m/s2[/tex]

In conclusion, the speed of the negative ball

[tex]v1=\sqrt{2ah}\\\\v1=\sqrt{2*9.67*2m}[/tex]

v1=6.21m/s

The  difference in v

V=v1-v2

V=6.31-6.21

V=0.10m/s

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#A

[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}[/tex]

[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}[/tex]

[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}[/tex]

[tex]\\ \rm\Rrightarrow H_{max}=20sin^237[/tex]

[tex]\\ \rm\Rrightarrow H_{max}=7.2m[/tex]

#B

[tex]\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}[/tex]

[tex]\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}[/tex]

[tex]\\ \rm\Rrightarrow R=40sin74[/tex]

[tex]\\ \rm\Rrightarrow R=38.5m[/tex]

#C

[tex]\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}[/tex]

[tex]\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}[/tex]

[tex]\\ \rm\Rrightarrow T=4sin37[/tex]

[tex]\\ \rm\Rrightarrow T=2.4s[/tex]

Now

[tex]\\ \rm\Rrightarrow v=u-gt[/tex]

[tex]\\ \rm\Rrightarrow v=20-10(2.4)[/tex]

[tex]\\ \rm\Rrightarrow v=20-24[/tex]

[tex]\\ \rm\Rrightarrow v=-4m/s[/tex]

The time needed to evaporate all the water in the kettle is  1.13 x 10³s.

The power is defined as the the ratio of amount of energy produced per unit time.

P =E/t

Energy is in the form of heat gained to evaporate all water in kettle.

E = Q =mL

So, P = mL/t

Substitute mass of water m = 0.5kg,  latent heat of vaporization of wate L = 2.26 x 10⁶ J/kg, power P= 10³ W, we have

10³ = (0.5 x 2.26 x 10⁶) /t

t = 1.13 x 10³s

Thus, the time needed to evaporate all the water is  1.13 x 10³s.

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Answer:

The rim is usually heated before fixing into the wooden wheel because on heating the iron rim, it expands first, then on cooling contracts and fits onto the wooden wheel.

Explanation: Hope it helps you :))))

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Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

[tex]F = \dfrac{kq_1q_2}{r^2}[/tex]

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

[tex]F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}[/tex]

Force due to Q₃ :

[tex]F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}[/tex]

8.3. The net force on the particle at Q₂ is the vector

[tex]\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}[/tex]

Its magnitude is

[tex]\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}[/tex]

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

[tex]\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}[/tex]

where we subtract 180° because [tex]\vec F[/tex] terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

The displacement of Edward in the westerly direction is determined as 180.27 km.

The displacement of Edward is calculated as follows;

R² = a² + b² - 2abcosθ

R² = 150² + 200² - 2(150 x 200) x cos60

R² = 32500

R = 180.27 km

Thus, the displacement of Edward in the westerly direction is determined as 180.27 km.

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Answer:

increases

Explanation:

The ball's kinetic energy is the energy it has due to its motion. When you throw the ball into the air, it gains kinetic energy from the force of your throw.

0.24 m  distance  through which the piston rises.

we have given the mass of piston+block = 135kg

2050 joule of heat flows into the gas

now;

According to first law thermodynamics,

dQ=du+dw

we have

2050=1730+MgS

Here M is mass, g is gravity, S is distance

now puttin gthe formula

MgS=320

S=320/1323

S= 0.24m

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Answer:

3 820 885 N

Explanation:

Gravitational equation

   F = G  m1 m2 / r^2    

         G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11   *   4.41 x 10^5  * 5.97 x 10^24  / ( 6.78x 10^6)^2

 = 3820885 .3 N

A. The ball's (vertical) velocity [tex]v[/tex] at time [tex]t[/tex] is

[tex]v(t) = 30\dfrac{\rm m}{\rm s} - gt[/tex]

so that after 4 seconds, the ball's speed is

[tex]|v(4\,\mathrm s)| = \left|30\dfrac{\rm m}{\rm s} - \left(10\dfrac{\rm m}{\mathrm s^2}\right) (4\,\mathrm s)\right| = \boxed{10\frac{\rm m}{\rm s}}[/tex]

(The velocity is -10 m/s, so the ball is falling back down at this point.)

B. At maximum height, the ball has zero velocity, so it takes

[tex]30\dfrac{\rm m}{\rm s} - gt = 0 \implies t = \dfrac{30\frac{\rm m}{\rm s}}g = \boxed{3\,\mathrm s}[/tex]

for the ball to reach this height.

C. The height of the ball [tex]y[/tex] at time [tex]t[/tex] is

[tex]y(t) = \left(30\dfrac{\rm m}{\rm s}\right) t - \dfrac g2 t^2[/tex]

The maximum height is attained by the ball at 3 seconds after it's thrown, so

[tex]y_{\rm max} = \left(30\dfrac{\rm m}{\rm s}\right) (3\,\mathrm s) - \dfrac{10\frac{\rm m}{\mathrm s^2}}2 (3\,\mathrm s)^2 = \boxed{45\,\mathrm m}[/tex]

D. The time it takes for the ball to reach its maximum height is half the time it spends in the air. So the total airtime is [tex]\boxed{6\,\mathrm s}[/tex].

Put another way: When the ball returns to the height from which it was thrown, its final velocity has the same magnitude as its initial velocity but points in the opposite direction. This is to say, after the total time the ball is in the air, it's final velocity will be -30 m/s. Then the total airtime is

[tex]30\dfrac{\rm m}{\rm s} - gt = -30\dfrac{\rm m}{\rm s} \implies t = \dfrac{60\frac{\rm m}{\rm s}}g = \boxed{6\,\mathrm s}[/tex]

Put yet another way: Solve [tex]y(t) = 0[/tex] for [tex]t[/tex]. I don't see a need to elaborate...

The coefficient of friction between the road and the car's tire is determined as 0.78.

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

μ = a/g

μ = (7.72)/9.8

μ = 0.78

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The image of the object is 8cm to the left of the lens (D)

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

From the formula

1÷v = 1÷f - 1÷u

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

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Answer:

B

Explanation:

Resistors in series add because there is only one path the current can take.

calculations

= R1 + R2

= 50 + 50

= 100

There are three basic food groups. The foods are classified into Go, Grow and Glow.

Go Foods are the type of food that provide fuel and help us 'go' and be active whole day. 'Go' foods are: bread, rice, pasta, cereals and potato.

Grow foods are the ones which makes us grow by giving muscle strength and purifies blood.

Glow foods are the one which make us glow by face and skin healthy.

Foods classified into the three groups:

Toast+ Eggs  + Spinach : GO -GROW-GLOW

oatmeal + nuts + berries : GO -GROW-GLOW

stir fried tofu + broccoli + rice : GROW-GLOW -GO

toast + apples + peanut butter: GO -GROW-GLOW

crackers+ hummus + cucumber  :GO -GROW-GLOW

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The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²

Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²

It is expressed thus;

α= Δω ÷ Δt

Where α = angular acceleration

Δw = change in angular velocity = velocity ÷ radius

Δ t = change in time

Using the formula:

α= Δω ÷ Δt

v = 5. 33m/s, r = 0. 33m and t = 6.27s

Substitute the values to get the angular velocity

Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s

Substitute the value of Δw into the equation

α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²

Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²

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