An apple is falling from its tree. While falling, its acceleration changes
but not direction. Explain.

Answer:

The SI unit of power is watt

Answer:

u=20 m/s, T=4s

Explanation:

Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2

From equation of motion 

v2=u2+2gs−u2=−2(10).20u=20m/s

and time t can be determined by the formula

t=gv−u=−10−20=2s

total time = 2× time of ascend=2×2=4s

it is helpful for you

Answer:

The distance is 22.45 cm.

Explanation:

Height of cylinder, H = 14 cm

depth of hole, h = 5 cm

The distance of landing of stream from the base of cylinder is

[tex]R = 2\sqrt{H(H-h)}\\\\R = 2\sqrt{14(14-5)}\\\\R = 22.45 cm[/tex]

Answer:

in which topic you need help

Answer:

[tex]T=1.54s[/tex]

Explanation:

From the question we are told that:

Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]

Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]

Therefore

Frequency of Motor 1 [tex]f_1=13.75[/tex]

Frequency of Motor 2  [tex]f_2= 13.1[/tex]

Generally the equation for Time Elapsed is mathematically given by

[tex]T=\frac{1}{df}[/tex]

Where

[tex]df=f_1-f_2[/tex]

[tex]df=13.75-13.1[/tex]

[tex]df=0.65Hz[/tex]

Therefore

[tex]T=\frac{1}{65}[/tex]

[tex]T=1.54s[/tex]

Answer:

s=50m

Explanation:

you can use the formula

s=ut+1/2at²

s=0t+1/2(1)10²

=1/2(100)

=50

I hope this helps

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )

f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )

hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )

we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[  ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 =  ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that

0² - (73 ft/s)² = 2 (-24 ft/s²) x

==>   x ≈ 111.0 ft

Answer:

2.3  ×  [tex]10^{-1}[/tex]  

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

Answer:

0.23×10⁴

Explanation:

kilogram to gram ÷ 1000

0.00023kg ÷ 1000

=0.23g

scientific notation=0.23×10⁴

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

[tex]U = 0.5 C V^2[/tex]

The capacitance is given by

[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]

Answer:

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

Explanation:

Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:

Al:

[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)

Jo:

[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)

Total mass:

[tex]m_{1} + m_{2} = 194\,kg[/tex]

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.

[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.

[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.

[tex]F[/tex] - Impact force between skaters, in newtons.

[tex]\Delta t[/tex] - Impact time, in seconds.

If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:

[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)

[tex]m_{2} \cdot 6.7 = F[/tex] (2b)

(2b) in (1b):

[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]

[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]

[tex]14.6\cdot m_{2} = 1532.6[/tex]

[tex]m_{2} = 104.973\,kg[/tex]

[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]

[tex]m_{1} = 89.027\,kg[/tex]

And the magnitude of the force is:

[tex]F = 6.7\cdot m_{2}[/tex]

[tex]F = 596.481\,N[/tex]

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

Answer:

[tex]h=10m[/tex]

Explanation:

From the question we are told that:

Area [tex]a=70 x 10^{-6}[/tex]

Force [tex]F=7N[/tex]

Generally the equation for Pressure is mathematically given by

Pressure = Force/Area

[tex]P=\frac{F}{A}[/tex]

[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]

[tex]P= 1*10^{5} Pa[/tex]

Generally the equation for Pressure is also mathematically given by

[tex]P=hpg[/tex]

Therefore

[tex]h=\frac{P}{hg}[/tex]

[tex]h=\frac{10000}{1000*9.8}[/tex]

[tex]h=10m[/tex]

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex]  Rounding to the correct number of sig fig's to simplify:

[tex]v=\sqrt{400+2.0*10^2}[/tex] to get

v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

Answer:

0.032 [tex]m/s^2[/tex]

Explanation:

Given :

Weight of the astronaut = 69 kg

Weight of the physics book = 28 kg

Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]

The force that is applied on the astronaut :

[tex]F=ma[/tex]

   [tex]$=69 \times 0.013$[/tex]

   = 0.897 N

Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N

Therefore, the acceleration of the physics book is given by :

[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]

[tex]$a = \frac{0.897}{28}$[/tex]

a = 0.032 [tex]m/s^2[/tex]

Hence, the acceleration of the physics book is  0.032 [tex]m/s^2[/tex].

Answer:

The acceleration of astronaut is 5.27 x 10^-3 m/s^2.

Explanation:

mass of astronaut, M = 69 kg

Mass of book, m = 28 kg

acceleration of book, a = 0.013 m/s^2

Let the acceleration of astronaut is A.

According to the Newton's third law, for every action there is an equal and opposite reaction.

So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.

M A = m a

69 x A = 28 x 0.013

A = 5.27 x 10^-3 m/s^2

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.

Answer:

Energy = .13 W / m^2      energy of incident energy

N = 3500 Watts / day       power needed

N = 3500 Watts  (3600 * 24 sec) = .0405 Watts/sec

The problem must mean that one needs 3.5 Kw-days

3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules

150 J/sec-m^2 * .13 = 19.5 J / sec-m^2      usable energy

In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2    usable energy received

Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2

One would need 180 m^2 of solar panels

That's quite a lot of energy

A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.

Answer:

b. The heavier one reaches the bottom first.

Answer:

B

Explanation:

The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.

Answer:

the required angle is 0.0834879⁰

Explanation:

Given  the data in the question;

slit separation; d = 1.75 mm = 1.75 × 10⁻³ m

wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m

wavelength λ₂ 510 nm = 510 × 10⁻⁹ m

Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;

tanθ = [tex]y_m[/tex] / D = mλ/d

since they both coincides;

tanθ₁ = tanθ₂

m₁λ₁/d = m₂λ₂/d

multiply both sides by d

so,

m₁/m₂ = λ₂/λ₁

we substitute

m₁/m₂ = 510 nm / 425 nm

m₁/m₂ = 510 nm / 425 nm

divide through by 85

m₁/m₂ = 6 / 5

hence m₁ and m₂ are 6 and 5

so, from the previous formula

tanθ₂ = m₂λ₂/d

we substitute

tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 0.00145714

θ₂ = tan⁻¹( 0.00145714 )

θ₂ = 0.0834879⁰

Therefore, the required angle is 0.0834879⁰

Answer:

1.6031 miles

Explanation:

Given the following data;

Speed = 344 m/s

Time = 7.5 seconds

To find how many miles away was the lighting strike;

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 344 * 7.5

Distance = 2580 meters

Next, we would have to convert the value of the distance travelled in meters to miles;

Conversion:

1609.344 metres = 1 mile

2580 meters = X mile

Cross-multiplying, we have;

X * 1609.344 = 2580

X = 2580/1609.344

X = 1.6031 miles

This question is incomplete, the complete question is;

A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.

(a) Determine the magnitude of the emf induced in the loop.

(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)

Answer:

a) the magnitude of the emf induced in the loop is 0.003 V

b) dA/dt = 0.002 m²/s

Explanation:

Area of the loop wire A = 0.015 m²

magnitude of the field is increasing dB/dt = 0.20 T/s

a)

Determine the magnitude of the emf induced in the loop.

V = A( dB/dt )

we substitute

V = 0.015 m² × 0.20 T/s

V = 0.003 V

Therefore,  the magnitude of the emf induced in the loop is 0.003 V

b)  the induced emf is;

V = B( dA/dt ) + A( dB/dt )

given that; induced emf is 0, B = 1.5

so we substitute

0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]

-[ 1.5T × ( dA/dt )] = 0.003 m²T/s

dA/dt = -[ 0.003 m²T/s / 1.5T ]

dA/dt = -0.002 m²/s

the negative shows that the area is decreasing

hence, dA/dt = 0.002 m²/s

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because any object at rest or in uniform motion has no speed or velocity

Answer:

[tex]_{18}^{39} } Ar[/tex]

Explanation:

The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.

[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]

Argon is a stable gas and is found in the group 8 on the periodic table of elements.

Answer:

Answer is below

Explanation:

39 18 Ar

Answer:

6.08

Explanation:

Given that,

Current, I = 0.8 A

Time, t = 76 h

We need to find the Ah rating of a battery. It can be calculated by taking the product of current and time. So,

Ah = (0.8)(76)

= 6.08 Ah

So, the Ah rating of the battery is 6.08.

Answer:

[tex]v_1=87.40m/s[/tex]

Explanation:

From the question we are told that:

Mass of arrow [tex]m=52g[/tex]

Mass of rock [tex]m_r=1.50kg[/tex]

Height [tex]h=0.47m[/tex]

Generally the equation for Velocity is mathematically given by

 [tex]v = \sqrt{(2gh)}[/tex]

 [tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]

 [tex]v= 3.035m/s[/tex]

Generally the equation for conservation of momentum is mathematically given by

 [tex]m_1v_1=m_2v_2[/tex]

 [tex]0.052kg * v = 1.5 * 3.03m/s[/tex]

 [tex]v_1=87.40m/s[/tex]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]

Calculating the common potential:

[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]

Calculating the charge after redistribution:

[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]        

[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J