need help pleaseee,question is in the pic​

Answer:

1.5kg

Explanation:

Given data

mass m1= 2.5kg

mass m2=??

velocity of mass one v1= 1.5m/s

velocity of mass two v2= 4.1m/s

common velocity after impact v= 2.5m/s

Let us apply the formula for the conservation of linear momentum for inelastic collision

The expression is given as

m1v1+ m2v2= v(m1+m2)

substitute

2.5*1.5+ m2*4.1= 2.5(2.5+m2)

3.75+4.1m2= 6.25+2.5m2

collect like terms

3.75-6.25= 2.5m2-4.1m2

-2.5= -1.6m2

divide both sides by -1.6

m2= -2.5/-1.6

m2= 1.5 kg

Hence the second mass is 1.5kg

(C)

Explanation:

[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]

[tex]= 5.3×10^{-18}\:\text{J}[/tex]

Answer:

It's D

Explanation:

It's from alvs

Answer: (15 - 0)/1.8 = 8. 33m/s^2

Explanation:

The acceleration of the racehorse is 8.33 m/s²

The given parameters;

initial velocity of the racehorse, u = 0

final velocity of the racehorse, v = 15  m/s

time of motion of the horse, t = 1.8 s

The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]

Thus, the acceleration of the racehorse is 8.33 m/s²

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Using the left-hand rule,

[tex](4\,\vec\imath+\vec k)\times(2\,\vec\imath+\vec\jmath-3\,\vec k) = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\4&0&1\\2&1&-3\end{vmatrix} = -\vec\imath+14\,\vec\jmath+4\,\vec k[/tex]

Then in the right-handed rectangular coordinates, the cross product is the negative of this,

[tex]\boxed{\vec\imath-14\,\vec\jmath-4\,\vec k}[/tex]

When the rock is suspended in the air, the net force on it is

∑ F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

∑ F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

∑ F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

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

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

Answer:

You should multiply 60 kg*9.8 and answer will come.

Hope this will help you.

Answer:

yes she is right you should multiple 60*9.8

have a great day God bless you

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

Answer:

hhhhhhhjjjkkllkcftkbgfjknhglncg

Answer:

momentum (m)=36.9N

velocity (v)=7.56m/s

now,

momentum (m)=m×v

36.9=m×7.56

36.9÷7.56=m

m=4.89kg

Answer:

x= 2*I1*d/(I1+I2) meter

Answer:

https://youtu.be/ymHHdoCGJOU

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......

Answer:

impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction. The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg⋅m/s).

It stands for energy=mass times the speed of light squared.

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation:

Answer:

a. Find the graph in the attachment

b. 720 kΩ

c. The ratio V/I gives us our resistance which is 720 kΩ

Explanation:

a) plot a graph of V against I.

To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.

b) Determine the wire's resistance , R.

The resistance of the wire is determined as the gradient of the graph.

R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)

Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA

R = (144 V - 72 V)/(0.2 - 0.1) mA

R = 72 V/0.1 mA

R = 72 V/(0.1 × 10⁻³ A)

R = 720 × 10³ V/A

R = 720 kΩ

c) State ohm's law and try to relate it your results.​

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.

Mathematically, V ∝ I

V = kI

V/I = k = R

Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.

Since V/I is constant, we thus verify Ohm's law.

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  

So, option B is the answer

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]

Using the formula

Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

Learn more about wavelength and frequency, here:

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Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because any object at rest or in uniform motion has no speed or velocity

Answer:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

Answer:

Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem

《OAmalaOHopeO》

Answer:

In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Explanation:

_Hope it helps you_