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A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?

Answers

Answer 1

Answer:

Answer:

x = 0.68 meters

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Speed of the car, v = 25 m/s

Spring constant of the spring, $k=2* 10^6\ N/m$

When the car hits the uncompressed horizontal ideal spring the kinetic energy of the car is converted to the potential energy of the spring. Let x is the maximum distance compressed by the spring such that,

$(1)/(2)mv^2=(1)/(2)kx^2$

$x=\sqrt{(mv^2)/(k)}$

$x=\sqrt{(1500* (25)^2)/(2* 10^6)}$

x = 0.68 meters

So, the spring is compressed by a distance of 0.68 meters. Hence, this is the required solution.

Answer 2

Answer:

Final answer:

The maximum distance the spring compresses when a 1500 kg car moving at 25 m/s hits it, given a spring constant of 2.0 × 10⁶N/m, is approximately 0.53 meters or 53 centimeters.

Explanation:

In this specific problem, we can apply the conservation of energy principle, where the initial kinetic energy of the car is converted into potential energy stored in the spring when the car comes to a stop. The formula for kinetic energy is K = 1/2 × m× v² and for potential energy stored in a spring is U = 1/2×k × x², where m = mass of the car, v = velocity of the car, k = spring constant, and x = maximum distance the spring is compressed.

By setting the kinetic energy equal to potential energy (since no energy is lost), we get 1/2 × m×v² = 1/2×k×x². Solving this equation for x (maximum compression of the spring), we obtain x = sqrt((m×v²)/k). Substituting the given values, x = sqrt((1500 kg× (25 m/s)²) / (2.0 × 10⁶ N/m)), which yields approximately 0.53 meters or 53 centimeters. Therefore, the maximum distance the spring compresses is 53 cm.

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When you walk at an average speed (constant speed, no acceleration) of 24 m/s in 94.1 secyou will cover a distance of__?
A train moves with a uniform velocity of 36km/hr 10sec. calculate the distance travelled
A student lifts their 75 N backpack 0.50 m onto their chair. How much work is done?
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.(a) Find the electric force on charge q2.F12 = ? mN(b) Find the electric force on q1.F21 = ? mN(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?

The normal is a line perpendicular to the reflecting surface at the point of incidence.

Answers

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

A disgruntled autoworker pushes a small foreign import offacliff with a height of h. the vehicle lands a distance away
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.

Answers

Answer:

v = a/√(2h/g) m/s

Explanation:

Lets say the distance away from the cliff is a.

then, a = v t

where v is velocity with which it was thrown and t is time taken to fall.

Using equations of motion, we can also say that

h=1/2gt^2

where h is the height of the cliff

Thus, t^2 = 2h/g and t = √(2h/g)

Thus, v = a/√(2h/g).

the vehicle was pushed off the cliff with the velocity , v = a/√(2h/g). m/s

A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop isa. counterclockwiseb. clockwisec. there is no current in the loop because magnetic field is uniformd. there is no current in the loop because magnetic field does not change

Answers

Complete Question

a. counterclockwise

b. clockwise

c. there is no current in the loop because magnetic field is uniform

d. there is no current in the loop because magnetic field does not change

Answer:

Option A is the correct answer

Explanation:

According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

$U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)$ (1)

Where:

$U_(e)$ - Electric potential energy, measured in joules.

$q_(A)$ , $q_(B)$ - Electric charges, measured in coulombs.

$r$ - Distance between charges, measured in meters.

$k$ - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that $U_(e) = -24\,J$ , $q_(A) = 45* 10^(-6)\,C+ q_(B)$ , $k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2))$ and $r = 0.29\,m$ , then the electric charge is:

$-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)$

$-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)$

$9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0$ (2)

Roots of the polynomial are found by Quadratic Formula:

$q_(B,1) = 1.327* 10^(-5)\,C$ , $q_(B,2) \approx -5.827* 10^(-5)\,C$

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Answers

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

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