A cannonball is shot from the top of a
47.8 m high hill at a speed of 53.7 m/s.
How fast is it going when it reaches
the plain below?

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:

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.

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

Answer:

four glorps

Explanation:

We know :

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

[tex]$\Rightarrow -1 \text{glorp} = 0 - \frac{g}{2} \times (1 g\text{ gleep})^2$[/tex]

[tex]$\Rightarrow 1 \text{ glorp}= \frac{g}{2} (1 \text{ gleep})^2$[/tex]   .............(i)

Now, t' = 2 gleep

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

  [tex]$=0+ \frac{-g}{2} (2 \text{ gleep})^2$[/tex]

  [tex]$=-\frac{4g}{2}(2 \text{ gleep})^2$[/tex]

 [tex]$=4\left[\frac{-g}{2} (\text{gleep})^2\right]$[/tex]

 = 4 (-1 gleep)     (From (i))

So, |y| = 4 glorp

Answer:

c

Explanation:

search it up

Answer:

Option A.

Explanation:

Because this is a lunar eclipse it normally happens 2 times a year.  This is a decently rare phenomenon due to the positions they have to be in to make a lunar eclipse.  Therefore, it is option A.

Answer:

5 x 10⁻⁵ T

Explanation:

Magnetic field at a point d distance away from a current carrying wire having current of I

B = 10⁻⁷ x 2I / d

Here magnetic field due to current of 20 A

B₁ = 10⁻⁷ x 2 x 20 / 10 x 10⁻²

= 4 x 10⁻⁵ T .

Here magnetic field due to current of 5 A

B₂ = 10⁻⁷ x 2 x 5 / 10 x 10⁻²

=  10⁻⁵ T .

These magnetic fields are acting in the same direction

Total magnetic field

B = B₁ + B₂

= 4 x 10⁻⁵ + 10⁻⁵ = 5 x 10⁻⁵ T .

Answer:

check out of phase

Explanation:

this is my answer

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:

B - As one variable increases the other decreases. (-1 to +1)     coefficient

If the output work of a simple machine equals the input work, the machine is said to have 100% efficiency.

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

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].

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:

yong may check po ayon yong sagot

Answer:

static friction=0.126

Explanation:

x-component:

Vx = Vcos(theta)

= (92.5 m)cos(32.0)

= 78.4 m

Answer:

-78.4

Explanation:

For acellus students

Answer:

F = 668.9N

Explanation:

Using the formula;

m1d1 = m2d2

m1 and m2 are the masses

d1 and d2 are the distances

Substitute;

15 * 2.5 = x * (55/100) [55cm was converted to metres]

37.5 = 0.55x

Swap

0.55x = 37.5

x = 37.5/0.55

x = 68.18kg

Since F = mg

F = 68.18 * 9.81

F = 668.9N

This gives the required force

Answer:

5*10^2

Explanation:

A p e x

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:

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:

It is example of Closed system

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

You really don't need any help on the question.  It's all right there in the picture.  What you need help with is the answer.

The number of times the same thing happens each second is called its "frequency".  The frequency of the dragonfly's flaps is 477 Hz.  (If you're close enough to the dragonfly, you can hear the wings flapping.  It sounds like a raspy tone with a frequency of 477 Hz.)

The "period" is just the length of time it takes to happen once.  That length of time is just  (1 / frequency) .

The dragonfly flaps its wings once every  (1 / 477 Hz) = 0.0021 second (C)

[tex]\huge{ \mathfrak{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Copper wires are used as connecting wires in circuits because they offer very less resistance.

__________________________________________________________

[tex]\mathrm{ ☠ \: TeeNForeveR \:☠ }[/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:

S = Vo t * 1/2 a t^2     equation for distance traveled

w = w0 t + 1/2 a t^2    equivalent circular equation where w equals omega

w = 5.98 * 27.5  + a (27.5)^2/ 2

a = (w2 - w1) / t = -5.98 / 27.5 = -.217

w = 5.98 * 27.5 - 1/2 * .217 * 27.5^2 = 82.4

Since 1 Rev = 2 pi  radians      

Rev = 82.4 / 2 * pi = 13.1 Rev

Answer:

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

Explanation:

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:

Properties of matter

Explanation:

All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.

Answer:

The pythagoream theorem

Explanation: