1 point
4. When mass is in kilograms and velocity is in meters per second then
momentum is in kilograms-meters per second.
True
Ο Ο
False

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:

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:

Option C. 4 Hz

Explanation:

To know the correct answer to the question given above, it is important we know the definition of frequency.

Frequency can simply be defined as the number of complete oscillations or circles made in one second.

Considering the diagram given above, the wave passes through the medium over a period of one second.

Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.

From the diagram given above,

The number of circles = 4

Thus,

The frequency is 4 Hz

Answer:

ook soooooo

Explanation:

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:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m  

Period:

The period is given as:

[tex]Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\[/tex]

Period = 10 s

Frequency:

The frequency is given as:

[tex]f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\[/tex]

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

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]

Explanation:

x-component:

Vx = Vcos(theta)

= (92.5 m)cos(32.0)

= 78.4 m

Answer:

-78.4

Explanation:

For acellus students

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:

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:

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:

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:

yong may check po ayon yong sagot

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.

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:To explian the exact calculation of this pruduct we must not try!

Answer:

check out of phase

Explanation:

this is my answer

The relative error in this weight measurement as percentage is

is the ratio of the absolute error of a measurement to the measurement being taken. In other words, this type of error is relative to the size of the item being measured.

Therefore,

[tex]\% relative error = \frac{smallest division}{mass of apples}*100\\\\\% relative error = \frac{2.1g}{1900}*100\\\\\% relative error = 0.1105\%[/tex]

For more information on relative error, visit

https://brainly.com/question/64815

Answer:

The relative error in this weight measurement is approximately 0.1105 %.

Explanation:

We know that,
Relative Error is the ratio of the absolute error of a measurement to the actual measurement . It can be mathematically represented as,

Relative Error  = (Measured Value - Actual Value)  / Actual Value * 100

In this case, the actual weight is 1.9 kg, and the smallest division of the spring scale is 2.1 g.

Actual Weight = 1.9  * 1000  = 1900  (As 1 kg = 1000g)

We know that the absolute error will be equal to the smallest division hence,

Measured Value - Actual Value = 2.1 g

By replacing this in the general formula we get,

Relative Error = (Measured Value - Actual Value) / Actual Value * 100

= [tex]\frac{2.1}{1900} *100[/tex] %

= [tex]\frac{2.1}{19}[/tex] %

≈ 0.1105 %

Therefore, the relative error in this weight measurement is approximately 0.1105 %.

For more information on relative error, visit

brainly.com/question/64815

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]

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:

[tex]f_{U}= 1019.72hz[/tex]

Explanation:

From the equation we are told that:

Velocity of French sub [tex]V_{U}= 43.00km/h[/tex]

Velocity of U.S. sub at [tex]V_{F}|=64.00 km//h[/tex]

French Wave Frequency  [tex]F_{F}=1000Hz[/tex]

Velocity of wave  [tex]V_{s}=5470 km/h[/tex]

Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by

Doppler effect

 [tex]f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }[/tex]

 [tex]f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}[/tex]

 [tex]f_{U}= 1019.72hz[/tex]

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:

optimism

Explanation:

just an idea

....

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)