What is tan for the given triangle?

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:

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

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 pickle bottle cap on dipping it in hot water expanded.

Explanation: Hope it helps you:))))

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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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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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If only balanced forces act on an object it will keep moving in the same speed and direction.

This is defined as the type of quantity which has both magnitude and direction.

\Having balanced forces will bring about equilibrium which is  why the obect will move in the same speed and direction.

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The initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.

The initial speed of the second ball is calculated as follows;

v = √2gh

v = √(2 x 9.8 x 45)

v = 29.7 m/s

t = √(2h/g)

t = √(2 x 5/9.8)

t = 1.01 s

vf = vi + gt

29.7 = vi + 9.8(1.01)

29.7 = vi + 9.9

vi = 19.8 m/s

Thus, the initial speed of the second ball if they are both reach the ground at the same time is 19.8 m/s.

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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 weather forecast for the locations includes:

London - 21°C°, Precipitation: 60%, Humidity: 79%, Wind: 11 km/h.

New York - 26°C°, Precipitation: 50%, Humidity: 76%, Wind: 19 km/h.

It should be noted that weather is the state of the atmosphere with respect to heat or cold, calm or storm, clearness or cloudiness.

Weather forecast is an analysis of the state of the weather in an area with an assessment of the likely developments.

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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

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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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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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.

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.

The main issue that is being discussed in the manuscript is the development of science.

Capitalism impacts on science because extremely narrow and specialized bodies of knowledge creating problems for scientists to bring revolution. The dominant ideas in the society can also limit the development of science.

So we can conclude that the main issue that is being discussed in the manuscript is the development of science.

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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 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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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

[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]

Answer:

Hindi ko Po alm sorry Po talaga

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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Explanation:

Advantages

It is a good conductor of heat and hence, can measure even high temperatures.

It gives results quickly – has a fast response time.

It expands linearly and less than alcohol and any other liquid.

It is more durable than alcohol thermometer because mercury does not evaporate easily.

The force (expressed in N), the egg experience upon hitting the ground is 3.5N.

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given is an 0.07 kg egg and toss if from a window 30 m above the ground. The egg reaches and maintains a falling speed of 0.5 m/sec. Upon reaching the ground, the egg rapidly decelerates to 0 m/sec over the course of only 0.01 seconds.

F = m x (Vf -Vi)/t

F = 0.07 x (0 - 0.5)/0.01

F = -3.5N

Thus, the magnitude of force on egg upon hitting the ground is 3.5N.

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Answer:

2

Explanation:edu

Answer:

1.9

Explanation:

Efficiency=mechanical advantage/velocity ratio×100

95=M.A/2×100

95=50M.A

M.A=95/50=1.9

The mechanical advantage of the inclined plane is approximately 2.1053, and the effort required to raise the load of 400 newtons is approximately 189.78 newtons.

To determine the mechanical advantage and effort required for the inclined plane, we can use the formula:

Mechanical Advantage (MA) = Velocity Ratio (VR) / Efficiency

Effort Required = Load / Mechanical Advantage

Given:

Velocity Ratio (VR) = 2

Efficiency = 95% = 0.95

Load = 400 newtons

Calculate the Mechanical Advantage (MA):

MA = VR / Efficiency

MA = 2 / 0.95

MA ≈ 2.1053

Calculate the Effort Required:

Effort Required = Load / MA

Effort Required = 400 / 2.1053

Effort Required ≈ 189.78 newtons

So, the mechanical advantage of the inclined plane is approximately 2.1053, and the effort required to raise the load of 400 newtons is approximately 189.78 newtons.

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The heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

Heat conduction is the method of transfer of heat by the direct contact of the two or more solid bodies.

Given is the thickness of tissue x = 3cm =-0.03m, Area of conduction A = 1.5 m², Change in temperature of skin ΔT =2°C, and coefficient of thermal for tissue without blood is k = 18 × 10⁻² Cal /m-hr- °C

The conduction heat rate in hr is

Q =kAΔT /x

Q =18 × 10⁻² x 1.5 x 2 /0.03

Q =18 Joule / hr

Thus, the heat transferred by conduction per hour, if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

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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