A wire of negligible mass is wrapped around the outer surface of the disk of mass M. If the disk is released from rest, determine its angular velocity at time t. Given: M = 2 kg , t = 3 s , r = 80 mm

The height that will illustrate the distance will be d = 6.36m

Based on the information given, the length of the vine will be:

L = ✓(16² + 27.7)²

L = 32m

The velocity of Jane when she reaches position B will be:

V = ✓2gh

V = ✓(2 × 9.8 × 4.3)

V = 9.18m/s

We will apply the conversation of momentum. This will be:

50 × 9.18 = (50 + 80)V1

V1 = 3.53m/s

Therefore, the height that will illustrate the distance will be:

31.36² + d² = 32²

d² = 32² - 31.36²

d = 6.36m

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Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \(\frac{sin \theta}{cos \theta}\) = sin θ

           sin θ = y / L

we substitute

          \(d \ \frac{y}{L}\) = m λ

with the initial data we look for the distance between the lines

           d = \(\frac{m \lambda \ L}{y}\)

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

we reduce nm

           λ = 396.7 nm

The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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Answer: P=30W

Explanation:

formula is p=w/t

p = power

w = work

t = elapsed time

input variables, solve then simplify.

The power of a lift that transfers 450 J of energy in 15 seconds is 30 watts.

Power is defined as the rate of doing work, i.e. the amount of work done per unit time.

Mathematically, it can be represented as follows:

Power = Work done / time taken

Therefore, the power of a lift that transfers 450 J of energy in 15 seconds can be calculated as follows:

Power = Work done / time taken= 450 J / 15 s= 30 W

Therefore, the power of the lift is 30 watts.

To explain further, we know that power is measured in watts (W), and it is the rate at which work is done or energy is transferred.

Here, we are given that the lift transfers 450 J of energy in 15 seconds.

We can find the power of the lift by dividing the amount of work done by the time taken to do it. By substituting the given values, we get the power of the lift as 30 W.

In simple terms, this means that the lift can transfer energy at a rate of 30 joules per second. This can also be interpreted as the lift can do 30 joules of work in one second.

Hence, we can conclude that the power of the lift is 30 watts.

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Explanation

Step 1

free body diagram

Step 2

now, let's analyze the forces

a) in y

Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0)

so, as the leg is in rest

let

m= 2.27 Kg , so

hence

b) now in x ( horizontally)

therefore, the answer is

I hope this helps you

Answer:

a. f < M < T

Explanation:

Let us take the right direction as positive.

Since Ted exerts a force T to the right his force is +T, Mario exerts a force M to the left, his force is -M. It is also given that Mario's force is half of Ted's force, so M = T/2. Finally, the frictional force , f is to the left, so it is -f. Let the net force be F and it is to the right since the box moves to the right.

So, +T - M - f = +F

Substituting M = T/2, we have

+ T -T/2 - f = F

+ T/2 - f = F

+T/2 = F + f

So, T = 2(F + f) and

M = T/2 = F + f

Since T = 2(F +f) = 2M, It follows that T > M

Also, since M = F + f, it follows that M > f

So, T > M > f ⇒ f < M < T

So, the answer is a.

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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Explanation:

6) newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

Answer:

6 newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

Explanation:

this is right pls mark as brainliest

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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The apparent weight of the astronaut of mass 65.1 kg moving with a speed of 33.9 m/s in 2.7 s is 811.797 N.

Weight is the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration.

To calculate the astronaut's apparent weight, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the apparent weight of the astronaut is 811.797 N.

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The time elapsed between the first splash and the second splash is approximately 0.69 seconds.

To calculate this, we consider the motion of two rocks thrown simultaneously from a bridge. Heather throws a rock straight down with a speed of 14 m/s, while Jerry throws a rock straight up with the same speed.

We use the equation for displacement in uniformly accelerated motion: s = ut + (1/2)at^2.

For Heather's rock, which is thrown downwards, the initial velocity (u) is positive and the acceleration (a) due to gravity is negative (-9.8 m/s^2). The displacement (s) is the height of the bridge (46 m).

Solving the equation, we find two possible values for the time (t): t ≈ -4.91 s and t ≈ 1.91 s.

Since time cannot be negative in this context, we discard the negative value and consider t ≈ 1.91 s as the time it takes for Heather's rock to hit the water.

For Jerry's rock, thrown upwards, we use the same equation with the same initial velocity and acceleration. The displacement is also the height of the bridge, but negative.

Solving the equation, we find t ≈ -5.68 s and t ≈ 1.22 s. Again, we discard the negative value and consider t ≈ 1.22 s as the time it takes for Jerry's rock to reach its maximum height before falling back down.

To find the time difference between the first and second splash, we subtract t ≈ 1.91 s (Heather's rock) from t ≈ 1.22 s (Jerry's rock). This gives us a time difference of approximately 0.69 seconds.

Therefore, the time elapsed between the first splash and the second splash is approximately 0.69 seconds.

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The amount of charge that will pass through the cross-sectional area of the conductor in 1 min is 192 C.

We can use the formula Q = I * t,

where;

Given that a charge of 6.4 C passes through a cross-sectional area of the conductor in 2 s, we can find the current using the formula:

I = Q / t = 6.4 C / 2 s = 3.2 A

So, the current through the conductor is 3.2 A.

To find the amount of charge that will pass through the cross-sectional area of the conductor in 1 min (60 s), we can use the same formula:

Q = I * t = 3.2 A * 60 s = 192 C

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Answer:

The water pressure on the upper pipe is 92.5 kPa.

Explanation:

Given that,

Pressure in lower pipe= 120 kPa

Speed of water in lower pipe= 1 m/s

Acceleration due to gravity = 10 m/s²

Density of water = 1000 kg/m³

Radius of lower pipe = 12 m

Radius of uppes pipe = 6 m

Height of upper pipe = 2 m

We need to calculate the velocity in upper pipe

Using continuity equation

\(A_{1}v_{1}=A_{2}v_{1}\)

\(\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}\)

\(v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}\)

Put the value into the formula

\(v_{2}=\dfrac{12^2\times1}{6^2}\)

\(v_{2}=4\ m/s\)

We need to calculate the water pressure on the upper pipe

Using bernoulli equation

\(P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}\)

Put the value into the formula

\(120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2\)

\(120500=P_{2}+28000\)

\(P_{2}=120500-28000\)

\(P_{2}=92500\ Pa\)

\(P_{2}=92.5\ kPa\)

Hence, The water pressure on the upper pipe is 92.5 kPa.

Answer:

it can melt orcan put them past their boiling point

Explanation:

Answer:

Explanation:

For an ideal spring over a frictionless horizontal surface, stored energy is only a function of the spring constant k and the distance of compression. The mass of the block doing the compressing is irrelevant

Energy stored in the first example is

Em = ½kd²

Energy stored in the second example is

E₂m = ½k(2d)² = 4(½kd²) = 4Em

So the second situation has four times as much stored spring potential energy as the first situation

4 Em is correct

Good job!

Answer:Sir haymo knows

Explanation:

he gave as homework

The acceleration of the Saturn is - (2πf)²x(t).

A motion in which the restoring force is directly proportional to the body's displacement from its mean position is known as a simple harmonic motion, or SHM. This restoring force always moves in the direction of the mean position. A particle moving in simple harmonic motion accelerates as a(t) = - ω² x (t). Here, ω denotes the particle's angular velocity.

The acceleration of the particle at any position is directly proportional to the displacement from the mean position in simple harmonic motion, which is an oscillatory motion.

Given that:

at any time, the angular position of Saturn is (2πf)t.

the displacement in SHM at that time t is given by x(t)=R cos(2πf)t.

Hence, speed  in SHM at that time t is given by

\(v(t) = \frac{dx(t)}{dt} =R \frac{d }{dt} cos(2\pi f)t = -(2\pi f) sin(2\pi f)t\)

it's acceleration is

\(a(t) = \frac{dv(t)}{dt} =-(2\pi f) R \frac{d }{dt} sin(2\pi f)t = -(2\pi f)^2 Rcos(2\pi f)t = -(2\pi f)^2 x(t)\)

hence, at any time t, the acceleration of the Saturn is - (2πf)²x(t).

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Answer:

Aerosol spray lubricants should never be sprayed toward your body

Explanation:

Aerosols are substances that are enclosed under high pressure and released by means of a fine spray propelled by a gas.

It is very dangerous to spray aerosols towards the body as some of them are capable of causing severe damage to the body and even possible death; hence the answer above.

Answer:

Explanation:

Area= pier^2

=( 1.88/2)^2= 0.0000000314m^2

Intensity= 0.0265/0.000000314

= 7962W/m^2

Pressure= 2*7962)/345= 5.13*10^-5pa

Answer:

the answer is weight

Explanation:

Answer:

Weight

Explanation: Just took the quiz

The bowling ball has a rotating kinetic energy of 8.573 J and a total velocity of 134.573 J.

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the energy transfer, is done on it by exerting a net force. The word "kinetic" derives from the Greek "kinesis," which means motion. Any direction can be used to move it. As can be seen, kinetic energy rises with increasing mass and/or speed, and it stays unchanged if an object slows down or accelerates up.

To calculate rotational kinetic energy:

Rotational kinetic energy = (1/2) * I * ω^2

where I is the intertia of solid

I = (2/5) * m * r^2, m is the mass and r is radius

Substituting the given values, we get:

I = (2/5) * 7 kg * (0.109 m)^2

I = 0.00265 kg * m^2

The angular velocity of the ball ω = v / r

let v is the linear velocity of the ball.

Substituting the given values, we get:

ω = 6 m/s / 0.109 m

ω = 55.046 rad/s

by substituting this values into formulae we get

Rotational KE= (1/2) * 0.00265 kg * m^2 * (55.046 rad/s)^2

Rotational KE = 8.573 J

Therefore, the rotational kinetic energy of the bowling ball is 8.573 J.

The translational kinetic energy can be calculated as:

Translational kinetic energy = (1/2) * m * v^2

Substituting the given values, we get:

Translational KE= (1/2) * 7 kg * (6 m/s)^2

Translational KE = 126 J

Therefore, the total KE of the ball is:

Total kinetic energy = Rotational kinetic energy + Translational kinetic energy

Total KE = 8.573 J + 126 J

Total kinetic energy = 134.573 J

Therefore, the total kinetic energy of the ball is 134.57

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Answer:

Sorry I dont know this answer sorry

Answer:

P = 490500 [Pa]

Explanation:

The pressure at the bottom of a vessel and even of a lake or sea can be calculated by means of the following hydrostatic equation.

\(P=Ro*g*h\)

where:

P = pressure [Pa] (units of pascal)

Ro = water density = 1000 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 50 [m]

Now replacing:

\(P=1000*9.81*50\\P=490500[Pa]\)

Answer:

Granite

Explanation:

Trust me I learned this 2years ago

Answer: 707.11m

Explanation:

since the well is at 45 degrees, we can use trig ratios to figure out the vertical depth of the well as u can see image attached.

then since we are looking for the vertical depth and we have information on the hypotenuse we can say

sin45= \(\frac{verticle height}{1000}\)

therefore, we can say.

1000sin(45) = vertical height

hence

vertical height = 707.11m

Answer:

B

Explanation:

Answer:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

Explanation:

We'll begin by calculating the resistance. This can be obtained as follow:

Power (P) = 60 W

Voltage (V) = 220 V

Resistance (R) =?

P = V²/R

60 = 220² / R

Cross multiply

60 × R = 220²

60 × R = 48400

Divide both side by 60

R = 48400 / 60

R ≈ 807 Ohm

1. Determination of the number of dry cells of 1.5 V required.

Voltage (V) = 220

Dry Cells = 1.5 V

Number of dry cells (n) =?

n = Voltage / Dry cells

n = 60 / 1.5

n = 40

2. Determination of the number of internal resistance of 1 ohm required.

Resistance (R) = 807 Ohm

Internal resistance (r) = 1 ohm

Number of internal resistance (n) =?

n = R/r

n = 807 / 1

n = 807

SUMMARY:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

The car's vertical position \(y\) at time \(t\) is

\(y = 63\,\mathrm m - \dfrac12 gt^2\)

since it starts 63 m above the ground, and after leaving the cliff it accelerates downward due to gravity.

Its horizontal position \(x\) is

\(x = \left(29\dfrac{\rm m}{\rm s}\right) t\)

since the car leaves the cliff horizontally at 29 m/s, and is not influenced by any other acceleration in this plane.

1. Solve for \(t\) such that \(y=0\).

\(63\,\mathrm m - \dfrac12 gt^2 = 0 \implies t = \sqrt{\dfrac{126\,\rm m}g} \approx \boxed{3.6}\,\rm s\)

2. Solve for \(x\) at this value of \(t\).

\(x = \left(29\dfrac{\rm m}{\rm s}\right) \sqrt{\dfrac{126\,\rm m}g} \approx 104\,\rm m \approx \boxed{100}\,\rm m\)