Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measured in the laboratory. What is the magnitudeof the velocity of one particle relative to the other

Answer:

= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s

Explanation:

Answer:

     λ = 5.75 10⁻⁷ mm

Explanation:

This is a slit interference exercise, we analyze each wavelength separately                    

            λ = 657 nm                                     indicate that the third dark pattern

          a sin θ = (m + ½) lam

          a sin θ = (3 + ½) 657 10⁻⁹

          a sin θ = 2299.5 10⁻⁹ nm

for the other wavelength in the same place we have m = 4 bright

          a sin θ = m lam

we substitute

           2299.5 10⁻⁹ = 4 λ

           λ = [tex]\frac{2299.5 \ 10^9 }{ 4}[/tex]

           λ = 5.75 10⁻⁷ mm

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

Answer: [tex]857\ m[/tex]

Explanation:

Given

Speed of David car [tex]v=30\ m/s[/tex]

Tina begins to accelerate [tex]2.1\ m/s^2[/tex] after David pass the tina

Suppose it took t time for tina to catch David

Distance traveled by David in t time

[tex]\Rightarrow s_d=30\times t[/tex]

Using the equation of motion to get the distance of Tina is

[tex]s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2[/tex]

now, [tex]s_d=s_t[/tex]

[tex]30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s[/tex]

Neglecting [tex]t=0[/tex]

Distance traveled by tina in [tex]28.57\ s[/tex] is

[tex]s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m[/tex]

Answer:

5*10^2

Explanation:

A p e x

Answer:

a. hydroelectric power plant

Answer:

Height of the door = 2m = 2000 cm

1 in = 2.54 cm

So 1 cm = 1/2.54 in

2000 cm = 200000/ 254

=

787.401574803

So no.c is correct

The door is  78.7 inch tall. Hence, option (A) is correct.

Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.

The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.

Given that: the height of the door is = 2 meter

= 2*100 centimeter

= 200 centimeter.

There are 2.54 centimeter in 1 inch.

Hence,  the height of the door is = 2 meter  = 200 centimeter

= (200/2.54) inch

= 78.7 inch.

The door is 78.7 inch tall.

Learn more about length here:

https://brainly.com/question/17139363

#SPJ2

Answer:

the wavelength of the sound in seawater is 90.1 m.

Explanation:

Given;

frequency of the sound, f = 17 Hz

speed of the sound in seawater, v = 1531 m/s

The wavelength of the wave is calculated as follows;

v = fλ

λ = v / f

where;

λ is the wavelength of the sound

λ = 1531 / 17

λ = 90.1 m

Therefore, the wavelength of the sound in seawater is 90.1 m.

Answer:

check out of phase

Explanation:

this is my answer

Answer:

ook soooooo

Explanation:

Answer:

speed = 0.8c

Explanation:

Given :

Distance from earth to the distant planet = 10 ly

Time taken by the astronauts for the entire journey = 25 years

The time taken to reach the planet is [tex]$t_1=\frac{25}{2}$[/tex]

                                                                  = 12.5 years

Therefore, speed of the starship can be calculated by :

[tex]$\text{Speed} = \frac{\text{distance}}{\text{time}}$[/tex]

[tex]$v=\frac{10 \times c \times 3.15 \times 10^7}{12.5 \times 3.15 \times 10^7}$[/tex]

  [tex]$=0.8c$[/tex]

Therefore the speed of the starship is 0.8c

Answer:

Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit

mark me brainliestt :))

Answer:

[tex](a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

[tex](b)\ F_{No} = 771.125N[/tex]

Explanation:

Given

[tex]d_D = 6000ft[/tex] ---- Altitude of container in Denver

[tex]A = 0.0155m^2[/tex] -- Surface Area of the container lid

[tex]P_D = 79000Pa[/tex] --- Air pressure in Denver

[tex]P_{No} = 100250Pa[/tex] --- Air pressure in New Orleans

See comment for complete question

Solving (a): The expression for [tex]F_{No[/tex]

Force is calculated as:

[tex]F = \triangle P * A[/tex]

The force in New Orleans is:

[tex]F_{No} = \triangle P * A[/tex]

Since the inside pressure is half the pressure at sea level, then:

[tex]\triangle P = P_{No} - \frac{P_{area}}{2}[/tex]

Where

[tex]P_{area} = 101000Pa[/tex] --- Standard Pressure

Recall that:

[tex]F_{No} = \triangle P * A[/tex]

This gives:

[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

Solving (b): The value of [tex]F_{No[/tex]

In (a), we have:

[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

Where

[tex]A = 0.0155m^2[/tex]

[tex]P_{No} = 100250Pa[/tex]

[tex]P_{area} = 101000Pa[/tex]

So, we have:

[tex]F_{No} = [100250 - \frac{101000}{2}] * 0.0155[/tex]

[tex]F_{No} = [100250 - 50500] * 0.0155[/tex]

[tex]F_{No} = 49750* 0.0155[/tex]

[tex]F_{No} = 771.125N[/tex]

So they give us this

V=IR

V= 1.8

I=0.4

R=?

So we insert the thing that we know.

1.8=0.4*R

We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.

So we have this.

Lastly we solve.

R=4.5ohms

The formula to find R is V=IR

V/I=R

So the resistance will be the Voltage divided by the Current

Answer:

Rain water carries sediment and then these accumulate on the bottom of ponds, lakes and wetlands. This accumulation build up over time and eventually, the water disappears (because they sink into the ground) and the area once covered with water becomes land.

Answer:

static friction=0.126

Answer:

The answer is "[tex]4659.2 \times 10^{-24} \ N[/tex]"

Explanation:

The magnetic field at ehe mid point of the coils is,

[tex]\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\[/tex]

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

[tex]\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\[/tex]

       [tex]=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\[/tex]

Calculating the force experienced through the protons:

[tex]F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N[/tex]

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

Answer:

[tex]175\ \text{N/C}[/tex]

Explanation:

[tex]E_1[/tex] = Initial electric field = 3.5 N/C

[tex]E_2[/tex] = Final electric field

[tex]r_1[/tex] = Initial distance = 10 m

[tex]r_2[/tex] = Final distance = 20 cm

Electric field is given by

[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]

So,

[tex]E\propto \dfrac{1}{r}[/tex]

[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]

The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].

Answer:

d = 1.19 x 10⁻⁴ m = 0.119 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe spacing = 32 mm = 0.032 m

L = slit to screen distance = 6 m

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

d = slit width = ?

Therefore,

[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]

d = 1.19 x 10⁻⁴ m = 0.119 mm

Answer:

1. F = 3400 N = 3.4 KN

2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3. v = 14.9 m/s

Explanation:

1.

First, we will calculate the acceleration of Pam by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance = 27.3 m

vf = final speed = 62 m/s

vi = initial speed = 0 m/s

Therefore,

[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]

Now, we will calculate the force by using Newton's Second Law of Motion:

F = ma

F = (48.3 kg)(70.4 m/s²)

F = 3400 N = 3.4 KN

2.

Final kinetic energy is given as:

[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]

[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3.

According to the law of conservation of energy:

[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]

where,

v = speed at bottom = ?

g = acceleration due to gravity = 9.81 m/s²

h = height at top = 11.3 m

Therefore,

[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]

v = 14.9 m/s

Answer: [tex]1.102\ J[/tex]

Explanation:

Given

Mass [tex]m=1.3\ kg[/tex]

Spring constant [tex]k=58\ N/m[/tex]

Compression in the spring [tex]x=19.5\ cm\ or\ 0.195\ m[/tex]

When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.

[tex]\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J[/tex]

The kinetic energy of the mass is 1.102 J.

Answer:

Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria

Explanation:

Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.

It is finished in mitochondria.

Rt = 3 ohms

Explanation:

Let R1 = 4-ohm resistor

R2 = 12-ohm resistor

For 2 resistors connected in parallel, the total resistance Rt is given by

1/Rt = 1/R1 + 1/R2

or

Rt = R1R2/(R1 + R2)

= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)

= 48 ohms^2/16 ohms

= 3 ohms

Answer:

Cold

Explanation:

Im pretty sure im sorry if I am wrong

Complete question is;

A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The

centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.

four times

twice

half

equal to

Answer:

Half

Explanation:

Formula for centripetal force is given as;

F = mv²/R

Where;

v is velocity

R is radius

Now, centripetal acceleration is given by;

a = v²/R

Since they both travel with the same velocity V and radius remains the same, we can say that;

F = ma

For the small car;

FS = ma

For the big car;

FL = 2ma

This means the force of the small car is half of that of the Large car

Thus;

FS = ½FL