need in hurry please

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = [tex]\sqrt {2g \ y_1}[/tex]

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

Answer:

I don't see anything on your question?

Answer:

[tex]F=78.3hz[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=12[/tex]

Length [tex]l=80cm=0.8m[/tex]

Linear density [tex]\mu= 6.0g[/tex]

Generally the equation for Frequency is mathematically given by

 [tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]

 [tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]

 [tex]F=78.3hz[/tex]

I suppose you meant to say the radius of the curve is 260 m, not mm?

There are 3 forces acting on the car as it makes the turn,

• its weight mg pulling it downward;

• the normal force exerted by the road pointing upward, also with magnitude mg since the car is in equilibrium in the vertical direction; and

• static friction keeping the car from skidding with magnitude µmg (since it's proportional to the normal force), pointing horizontally toward the center of the curve.

By Newton's second law, the net force on the car acting in the horizontal direction is

F = ma   =>   µmg = ma   =>   a = µg

where a is the car's radial acceleration given by

a = v ^2 / R

with v = the car's tangential speed and R = radius of the curve. At the start, the car's radial acceleration is

a = (32 m/s)^2 / (260 m) ≈ 3.94 m/s^2

(a) If µ were reduced by a factor of 2, then the radial acceleration would also be halved:

1/2 a = 1/2 µg

Then the car can have a maximum speed v of

1/2 a = v ^2 / R   =>   v = √(aR/2) = √((3.94 m/s^2) (260 m) / 2) ≈ 22.6 m/s

(b) If µ were increased by a factor of 2, then the acceleration would also get doubled. Then the maximum speed v would be

2a = v ^2 / R   =>   v = √(2aR) = √(2 (3.94 m/s^2) (260 m)) ≈ 45.3 m/s

Answer:

We want to solve the sum:

6.74*10⁴ + 8.95*10⁴

first, we take the common factor 10⁴ out, so we get:

(6.74 + 8.95)*10⁴

Now we solve the sum:

(15.66)*10⁴

Now we want to rewrite it in exponential form, wo we can rewrite it as:

(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵

k = 5.

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

Answer: 0

Explanation: Trust

The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:

F = q * v * B * sin(θ)

Where:

F is the magnetic force,

q is the charge of the particle (in this case, 4.88 x 10^(-6) C),

v is the velocity of the particle (in this case, 265 m/s),

B is the magnetic field strength (in this case, 0.0579 T),

θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).

Plugging in the values:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)

Since sin(90°) is equal to 1, the equation simplifies to:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1

Calculating the value:

F = 6.47 x 10^(-4) N

Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

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The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.

The correct answer is option E.

To calculate the energy of a photon, we can use the equation:

E = (hc) / λ

where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.

Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:

E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)

E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)

E = 3.3695 x [tex]10^-^1[/tex] eV

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The question probable may be:

The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

E)  0.337 eV

Answer:

Explanation:

a)

Given that:

V = 25 mi/hr

To ft/sec, we have:

[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{110}{3} ft/s[/tex]

[tex]\rho = 0.075 \ lb/ft^3[/tex]

[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]

[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]

[tex]C_d = 0.28[/tex]

A = 25ft²

Recall that:

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]

[tex]F_d =10.967 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]

For 70 miles per hour, we have:

[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{308}{3} ft/s[/tex]

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]

[tex]F_d =85.99 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]

Answer:

The velocity after 1.5 s is 22.7 m/s downwards.

Explanation:

Initial velocity = - 8 m/s

acceleration, a = - 9.8 m/s2

time, t = 1.5 s

Use first equation of motion

v = u + at

v = - 8 - 9.8 x 1.5

v = - 8 - 14.7

v = - 22.7 m/s  

Thus, the velocity after 1.5 s is 22.7 m/s downwards.

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

Answer:

Parallel combination.

for the long life cells we have to connect them in parallel combination

hope it is helpful to you

Answer:

V = 2.58 m/s

Explanation:

Below is the calculation:

Given the weight of Erica = 37 kg

The weight of Danny = 45 kg

Danny's speed to move upward = 4.7 m/s

Use below formula to find the answer.

m1 * u1 = (m1+m2) * V

V = m1*u1 / (m1+m2)

Here, m1 = 45

u1 = 4.7

m1 = 45

m2 = 37

Now plug the values in formula:

V = m1*u1 / (m1+m2)

V = (45*4.7)/(45+37)

V = 2.58 m/s

Explanation:

9420 km/hr is the correct answer

Hope this helps...☺

Answer:

b)      v_y = 4.57 m / s

a)      vₓ = 4.43 m / s

Explanation:

This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity

          v = v₀ + a t

          v = 0 + a t

          v = 186  0.0342

          v = 6.36 m / s

let's use trigonometry to decompose this velocity

           sin 45.9 = v_y / v

           cos 45.9 = vₓ / v

           v_y = v sin 45.9

           vₓ = v cos 45.9

           v_y = 6.36 sin 45.9

           vₓ = 6.36 cos 45.9

           v_y = 4.57 m / s

           vₓ = 4.43 m / s

ₓ

Answer: The magnification of the magnifying glass is -2.9

Explanation:

The equation for power is given as:

[tex]P=\frac{1}{f}[/tex]

where,

P = Power = 10 D

f = focal length

Putting values in above equation, we get:

[tex]f=\frac{1}{10}=0.1m=10 cm[/tex]            (Conversion factor: 1 m = 100 cm)

The equation for lens formula follows:

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]

where,

v = image distance

u = Object distance = 6.55 cm

Putting values in above equation, we get:

[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]

Magnification (m) can be written as:

[tex]m=\frac{-v}{u}[/tex]

Putting values in above equation, we get:

[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]

Hence, the magnification of the magnifying glass is -2.9

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

Answer:

[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]

Explanation:

Given that,

The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]

A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]

Put all the values,

[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]

So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.  

Answer:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

Explanation:

The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.

The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]

From the given options, the second option is the correct answer.

[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]

Answer:

because it is helpful to human beings I think

Answer:

120 Newton

Explanation:

Given the following data;

Mass = 12 kg

Angle = 4°

We know that acceleration due to gravity is equal to 10 m/s

To find the minimum force to stop the block from sliding;

Force = mgCos(d)

Where;

m is the mass of an object.

g is the acceleration due to gravity.

d is the angle of inclination (theta).

Substituting into the formula we have;

F = 12*10*Cos(4°)

F = 120 * 0.9976

F = 119.71 ≈ 120 Newton

Answer:

Josiah Spiers in 1867 was when scripture union was founded

Answer:

Answer:

0.3333

Explanation:

Acceleration = change in velocity/time

a = 20 m/s / 60 m/s

a = 0.3333 m/s^2

k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.

When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.

For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.

In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.

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