A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put ​

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

Answer:

c). Easier setup

Explanation:

As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

Answer:

b. approaches zero.

Explanation:

The phenomenon is known as length contraction.

Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.

let the length of the train = L

Let the length observed when the train is in motion = L₀

Apply Einstein's special theory of relativity;

[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]

from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)

As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L).  (L₀ < L)

Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero.  (L₀ = 0)

Thus, the length of this object, as measured by a stationary observer approaches zero

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

Answer:

The initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

Given that,

Angular displacement, [tex]\theta=12\ rad[/tex]

Final angular speed, [tex]\omega_f=17\ rad/s[/tex]

The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]

We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,

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

Put all the values,

[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]

So, the initial angular speed of the CD is equal to 14.73 rad/s.

Explanation:

pls write the full question

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

Answer:

no, all substances doesnot conduct heat

Answer:

No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.

The bad conductors of heat are water, air, plastic, wood, etc.

Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

Answer:

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension

The string will be 108 cm long when it is stretched to the point where it becomes plastic.

Elasticity in physics and materials science refers to a body's capacity to withstand a force that causes distortion and to recover its original dimensions once the force has been withdrawn.

When sufficient loads are applied, solid objects will deform; if the material is elastic, the object will return to its original size and shape after the weights have been removed. Unlike plasticity, which prevents this from happening and causes the item to stay deformed,

Given parameters:

The elastic extensibility of a piece of string is 0.08.

The string is 100 cm long.

Hence,  it becomes plastic, after  it is stretched up to = 100 × 0.08 cm = 8 cm.  The string will be 108 cm long.

Learn more about elasticity here:

https://brainly.com/question/28790459

#SPJ5

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

Move towards the wall.

Explanation:

When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.

As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

[tex]\omega=109.67\ rad/s[/tex]

Explanation:

Given that,

The speed of the ball, u = 34 m/s

The ball is 0.310 m from the elbow joint.

We need to find the angular velocity (in rad/s) of the forearm.

We know that,

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]

So, the required angular velocity of the forearm is 109.67 rad/s.