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The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer 1

Answer:

Answer:

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K = 50N/m

New length = 32cm

Unknown

Force applied = ?

Solution:

The force applied on a spring can be derived using the expression below;

Force = KE

k is the spring constant

E is the extension

extension = new length - original length

extension = 32cm - 22cm = 10cm

convert the extension from cm to m;

100cm = 1m;

10cm will give 0.1m

So;

Force = 50N/m x 0.1m = 5N

Answer 2

Answer:

Final answer:

To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.

Explanation:

The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).

The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.

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A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up with your arms flat against your sides, and when you are standing straight up holding your arms straight out to your sides. Estimate the ratio of the moment of inertia with your arms straight out to the moment of inertia with your arms flat against your sides. (Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men's clothing stores, a man's average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches, from which an average circumference can be calculated. We'll also assume that about 20% of your body's mass is in your two arms, and that each has a length L = 1 m, so that each arm has a mass of about m = 8 kg.)

Answers

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall building. Assume air resistance is negligible, and g = 9.8 m/s2. What is the magnitude of the horizontal displacement of the rock?

Answers

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

$v_(x)=u\cos\theta$

Put the value into the formula

$v_(x)=11.5*\cos50$

$v_(x)=7.39\ m/s$

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Final answer:

The question is about determining the horizontal displacement of a projectile based on the given initial speed and projection angle and the height of the launch. This can be calculated using the equations of motion, specifically those pertaining to projectile motion.

Explanation:

In this problem, we're dealing with projectile motion. The stone being thrown is the projectile in this case. The horizontal displacement, also known as range, of a projectile can be defined using the formula: range = (initial speed * time of flight) * cosθ, where θ is the angle of projection. The initial speed is given as 11.5 m/s and the angle as 50 degrees. Now, we need to calculate the time of flight. This can be found by the formula: time of flight = (2 * initial speed * sinθ) / g. Considering g, the acceleration due to gravity, as 9.8 m/s², we can find the time of flight and thus calculate the range. Always remember that while the vertical motion of a projectile is affected by gravity, the horizontal motion remains constant.

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An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

Answers

Complete Question

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = (1)/(2)mv^2$

Making v the subject

$v = \sqrt{[(2 \Delta V * q )/(m)] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $(mv^2)/(r)$ ] and this is mathematically represented as

$Bqv = (mv^2)/(r)$

making B the subject

$B = (mv)/(rq)$

r is the radius with a value = 5.4cm = $= (5.4)/(100) = 5.4*10^(-2) m$

Substituting values

$B = (9.1039 *10^(-31) * 4.556 *10^7)/(5.4*10^-2 * 1.60218*10^(-19))$

$= 0.0048 T$

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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Enter your answer in the provided box. In water conservation, chemists spread a thin film of certain inert materials over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 32.0 m2 in area. Assuming that oil forms a monolayer (that is, a layer that is only one molecule thick) estimate the length of each oil molecule in nanometers. Assume that oil molecules are roughly cubic. (1 nm = 1 × 10−9 m)

Answers

Answer:

≅3.2 nm

Explanation:

Using the converter units as know for this case that:

1 ml is 1 cubic centimeter ⇒ 0.1 ml is 0.1 cubic centimeters

32.0 m² so :

32.0 m² *100 *100 cm² ⇒ 0.1 / ( 32.0 * 100 *100 ) = 100,000,000 * 0.1 / (32.0 * 100 * 100 ) nm

v = 100/32.0 nm = 3.125 nm thick.

v ≅3.2 nm

As oil is one molecule thick and the molecules are cubic, length of each oil is 3.2 nm

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(550-450)/(550)*3*10^(8)\n v=5.4545*10^(7)m/s$

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

$=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s$

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