* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)​

Answer:

 #_pile = 12 celdas

Explanation:

Lead acid sulfur batteries generate each cell a potential of 2 volts. By colonato to reach the voltage of 24 volts

        #_pile = 24/2

       #_pile = 12 cledas

serially connected

(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that

∑ F = s - mg = 0   ==>   s = mg = 617.4 N

If the spring is compressed 0.015 m, then the spring constant k is such that

617.4 N = k (0.015 m)   ==>   k = 41,160 N/m ≈ 41 kN/m

(2) The total mass of the passengers is

24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg

so that if everyone is at rest, the spring is compressed a distance x such that

kx = (1717 kg) g   ==>   x ≈ 0.41 m

Answer:

the study of the past

Explanation:

dogs and cats

Hydrogen bomb is more powerful than atom bomb

Answer:

The Michelson-Morley was designed to detect the motion of the earth through the ether.

No such relation was found and the speed of light is assumed to be the same in all reference frames.

The Michelson-Morley experiment was designed to measure: A. the velocity of the Earth relative to the ether.

Michelson-Morley experiment is an experiment which was first performed in Germany by the American physicist named, Albert Abraham Michelson between 1880 to 1881.

However, the experiment was later modified and refined by Michelson and Edward W. in 1887.

The main purpose of the Michelson-Morley experiment was to measure the velocity of planet Earth relative to the luminiferous ether, which is a medium in space that is hypothetically said to carry light waves.

In conclusion, the Michelson-Morley experiment was designed to measure the velocity of the Earth relative to the hypothetical luminiferous ether.

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Answer:

100 Joules

Explanation:

Applying,

W = mgh................... Equation 1

Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.

From the question,

Given: mg = 10 newtons, h = 10 meters.

Substitute these values into equation 1

W = 10×10

W = 100 Joules.

Hence the amount of workdone is 100 Joules

Answer:

a)  [tex]v=206.896m/s[/tex]

b)  [tex]L=6.749m[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=46Hz[/tex]

Length [tex]l=579m[/tex]

Total Mass [tex]T=4.3kg[/tex]

Tension [tex]T=3423[/tex]

a)

Generally the equation for velocity is mathematically given by

[tex]v=\sqrt{\frac{T}{\rho}}[/tex]

Where

[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]

Therefore

[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]

[tex]v=206.896m/s[/tex]

b)

Generally the equation for length of string is mathematically given by

[tex]L=\frac{3\lambda}{2}[/tex]

Where

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{206.89}{46}[/tex]

[tex]\lambda=4.498[/tex]

Therefore

[tex]L=\frac{3*4.498}{2}[/tex]

[tex]L=6.749m[/tex]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Answer:

x = 68.3 m

Explanation:

tan 32.8 = x / 106

Answer:

a)  [tex]F_g=1.5*10^9Ibf[/tex]

b)  [tex]F_t=12490Ibf/ft^2[/tex]

     [tex]F_b=0[/tex]

Explanation:

From the question we are told that:

Height [tex]h=200ft[/tex]

Width [tex]w=1200ft[/tex]

a)

Generally the equation for Dam's Hydro static force is mathematically given by

[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]

Where

[tex]\rho=Density\ of\ water[/tex]

[tex]\rho=62.4Ibm/ft^3[/tex]

Therefore

[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]

[tex]F_g=1.5*10^9Ibf[/tex]

b)

Generally the equation for Dam's Force per unit area is mathematically given by

[tex]F=\rho*g*h[/tex]

For Top

[tex]F_t=\rho*g*h[/tex]

[tex]F_t=62.4*32.2*200[/tex]

[tex]F_t=12490Ibf/ft^2[/tex]

For bottom

[tex]Here \\H=0 zero[/tex]

Therefore

[tex]F_b=0[/tex]

The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]

The force per unit area is the pressure exerted near the top of the dam.

[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]

The pressure near the bottom is zero, become the vertical height is zero.

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Answer:

5.4uC

Explanation:

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.

11.54 minutes

Explanation:

The decay rate equation is given by

[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]

where [tex]\lambda[/tex] is the half-life. We can rewrite this as

[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]

Taking the natural logarithm of both sides, we get

[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]

Solving for [tex]\lambda[/tex],

[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]

[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]

[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]

Answer: They are equal in magnitude.

- The signs of the two changes in potential energy are opposite

Explanation:

When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.

Also, it should be noted that the signs of the two changes in potential energy are opposite.

Hi,

The valves in the heart open and close to move blood in between heart chambers.

Answer:

d) one direction

Answer:

to the right

Explanation:

if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.

Answer:

The answer is 312.5j

Explanation:

The kinetic energy (KE):

KE=1/2*m*v^2

M= mass of the object

v= velocity of the object

We have;

m=25g

v=5m/s

KE=1/2*25g*5^2m/s

KE =312.5j

Answer:

chk photo

Explanation:

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

[tex] K = \frac{1}{2}mv^{2} [/tex]

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

Answer:

[tex]R=0.023m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.16*10^{-26}[/tex]

Potential difference [tex]V=523V[/tex]

Magnitude [tex]m=0.370 T[/tex]

Generally the equation for Velocity is mathematically given by

[tex]\frac{1}{2}mv^2=ev[/tex]

[tex]v=\frac{2ev}{m}[/tex]

[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]

[tex]v=12.22*10^4m/s[/tex]

Generally the equation for Force is mathematically given by

[tex]F=qvBsin \theta[/tex]

Where

[tex]qVB=m\frac{v^2}{R}[/tex]

[tex]F=m\frac{v^2}{R}sin\theta[/tex]

Therefore

[tex]R=\frac{mv}{qB sin \theta}[/tex]

[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]

[tex]R=0.023m[/tex]

Answer:

1.25

Explanation:

Answer:

yes there will be change in its density

Answer:

The depth of the resulting stream is 3.8 meters.

Explanation:

Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:

[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)

All volume flows are measured in cubic meters per second.

Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:

[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]

[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)

[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.

[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.

[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.

[tex]w_{3}[/tex] - Width of the resulting stream, in meters.

[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.

If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:

[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]

[tex]h_{3} = 3.8\,m[/tex]

The depth of the resulting stream is 3.8 meters.

Answer:

Option (a) is correct.

Explanation:

The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.

When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.

Answer:poop

Explanation:

poop

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D = D_A = D_B[/tex]

where;

length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]

[tex]\mathbf{Q_1 = 2Q}[/tex]

The flow rate in pipe B is 2Q of the flow rate of the pipe A

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D=D_A=D_B[/tex]

where;

length of the first pipe A  [tex]L_A=L[/tex] and the length of the second pipe B  

[tex]L_B=2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]

[tex]Q_1=2Q[/tex]

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

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Answer:

truee

Explanation:

Answer:

to prevent friction on the surfaces

Answer:

Explanation:

Oiling reduces friction between parts with relative motion between them.

Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.

Oil also becomes oxidized which reduces the oil's ability to decrease friction.

Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.