Vector approach in solving forces in 3d
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Answer:

PE (relative to water) = M g h

PE = 1200 kg * 9.8 m/s^2 * 24 m = 2.82E5 Joules

KE = PE when vehicle strikes water

1/2 M V^2 = 2.82E5

V = (2.82E5 * 2 / 1200)^1/2 = 21.7 m/s

Check:

M g h = 1/2 M V^2

V = (2 g h)^1/2 = (2 * 9.8 * 24)^1/2 = 21.7 m/s

Answer:

flammable  liquids

Explanation:when it catches on fire your going to get burned.hope this helps.

build a fence.....................

Answer:

proton is positively charged changechar

Explanation:

[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]

[tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]

[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]

Substitute the given parameters and solve for the angular speed;

[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

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The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.

The distance traveled by each object is determined from the prouduct of velocity and time of motion.

d = vt

where;

Thus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

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Answer:

Find the heavy processes of CO2 and methane.

Explanation:

Step #1 Find a very reliable energy source.

Step #2 Find the heavy processes of CO2 and methane like in many factories and places with high carbon emissions.

Step #3 Shut down each of the heavy processes even 10 per day would do a lot.

Step #4 Use the renewable energy source as a replacement for the attempts to generate energy using carbon emissions.

Answer:

(B)

Explanation:

Time = change of velocity ÷ acceleration

= (6-0) ÷ 5

= 1.2

1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.

Hence, the correct option is D.

To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:

time = change in velocity / acceleration

Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:

time = 6 m/s / 5 m/s²

time = 1.2 seconds

Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.

Hence, the correct option is D.

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Hi there!

We can use the angular equivalent of the following kinematic equation:

[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]

ωf = final angular velocity (1.9 rad/s)

ωi = initial angular velocity (11.3 rad/s)

α = angular acceleration (? rad/s²)

θ = angular displacement (17.95 rad)

We can rearrange the equation to solve for angular acceleration.

[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta\\\\\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}[/tex]

Plug in the given values and solve.

[tex]\alpha = \frac{1.9^2 - 11.3^2}{2(17.95)} = \boxed{3.456 \frac{rad}{s}}[/tex]

Answer:

See below

Explanation:

F = C q1 q2 /r^2

    8.988 x 10^9   *   (1.60217 x 10^-19)^2 / ( 5.29 x 10^-11)^2 =

.00000008244 N

82.4   nano N

Answer:

who can apply for the award

The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.

Power can be deifned as the rate at which work is done.

To calculate the power delivered to the truck, we use the formula below.

Formula:

Where:

From the question,

Given:

Subsitute these values into equation 1

Hence, The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.

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Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.

Given the following data:

First of all, we would determine the initial angular speed of the neutron star as follows:

[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]

Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:

[tex]I=\frac{2}{5} mr^2[/tex]

Where:

In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:

[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]

Final angular speed = 5,669 ≈ 6,000 rev/s.

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Answer:

1) asking a question about something you observe, 2) doing background research to learn what is already known about the topic, 3) constructing a hypothesis, 4) experimenting to test the hypothesis

Answer:

1) make an observation that describes a problem - see something you can fix or improve and try to describe the problem like which type of vegetables does rabbit like

2) create a hypothesis - What do you think will happen - If the rabbit eat the lettuce than .........

3) test the hypothesis - Do your experiment with the variables and follow the procedures

4) draw conclusions and refine the hypothesis - see if your hypothesis was correct - In conclusion my hypothesis was not correct because.......

Answer:

The circular turning of roads

Explanation:

please mark brainliest

Answer:

Explanation:

The circular turning of roads

Driving on Curves

For a class of 10 students taking an exam has a power output per student of about 200W, the temperature is mathematically given as

dT=25C

Generally, the equation for  the temperature is mathematically given as

dT=Q/mc

Where

Q=Pt

Q=2000*3600

Q=7200000J

And m

m=pv

m=1.3*270

m=351kg

Therefore

dT=7200000J/351*837

dT=25C

In conclusion, The temperature of the room at the end of 1.0 h

dT=25C

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The percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.

It should be noted that energy transformation simply means the process of changing energy from one form to another form.

In this case, the percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.

Also, the mass and height should be adjusted in such a way that the design of the system leads to GPE and KE values that are as close as possible.

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The experiment's percent difference between GPE and KE shows that the efficiency of energy transfer increases with percentage.

Thus, It should be emphasized that the term "energy transformation" simply refers to the act of altering energy's form.

The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.

The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.

Thus, Energy is capable of changing its forms. For instance, your body stores chemical energy from the food you eat until you can use it as kinetic energy when working or playing.

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(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]

[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]

The frequency of the sound heard is determined by applying Doppler effect.

[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]

where;

[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]

[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]

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Hi there!

We can begin by using Coulomb's Law:

[tex]E = \frac{kq}{r^2}[/tex]

k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)

E = Electric field strength (N/C)
r = distance from point (m)

q = charge (C)

Since this is a continuous charge, we must use calculus.

We can express this as the following:
[tex]q = \lambda L[/tex]

λ = Linear charge density (C/m)

L = Length of rod (m)

Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]

Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]

However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]

Integrate. For a semicircle, the bounds will be from -π/2 to π/2.

[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]

We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]

Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]

**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.

Answer:

True

Explanation:

Answer:

4

Explanation:

We know that :

Solving

The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

The linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.

The linear mass density of the rope is calculated as follows;

μ = m/L

μ = 0.6 kg / 4 m

μ = 0.15 kg/m

Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

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Answer:

its counting by 4 the multi 4,8,12,16

it has no energy because its being held still

it has motion energy because it will fall if let go

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .

For calculating the Force , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]

[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]

Hence :-

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