PHYSICS COLLEGE

Make the following conversion.56.32 kL = _____ L

0.056320
0.56320
5,632
56,320

Answers

Answer 1
Answer:

Answer:

The answer would be D 56,320

Explanation:

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A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Answers

The amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

What is potential energy?

Potential energy is the energy which body posses because of its position.

The potential energy of a body is given as,

PE=mgh

Here, (m) is the mass of the body, (g) is the gravitational force and (h) is the height of the body.

The energy stored in the spring is the sum of all the potential energy, kinetic energy and the energy dissipated due to friction. Therefore, it can be given as,

E=mgh+(1)/(2)mv^2+\mu mgd\cos\theta


Here, the mass of the wooden block is 1.05 kg . Angle of inclination is  35.0 degrees (point A). The distance from point B is 4.90m up the incline from A.

The speed of the block is 5.10 m/s and the coefficient of kinetic friction between the block and incline is 0.55. Therefore, put the values in the above formula as,

E=1.05(9.81)(4.9\sin(35))+(1)/(2)(1.05)(5.1)^2+(0.55)(1.05)(9.8)(4.9)\cos(35)\nE=65.3\rm J

Hence, the amount of potential energy that was initially stored in the spring due to the wooden block is 65.3 joules.

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Answer:

Explanation:

energy stored in spring initially

= kinetic + potential energy of block + energy dissipated by friction

= 1/2 mv² + mgh + μ mgcosθ x  d

m is mass , v is velocity at top position , h is vertical height , μ is coefficient of friction ,θ is angle of inclination of plane

= m (1/2 v² + gh + μ gcosθ x  d )

= 1.05 ( .5 x 5.1² + 9.8 x 4.9 sin35 + .55 x 9.8 cos35 x 4.9 )

= 1.05 ( 13.005 + 27.543 + 21.635)

= 65.3 J .

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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A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, concentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?A) a darkspotB) a bright spot thatis some color other than redC) a bright redspotD) a rainbow of color

Answers

Answer:

After passing through the glass plate, the red light disperses and meets at point.

The convex lens has two refracting surfaces, and convex kens is called as converging lens. So, at the exact center of the lens, one observes a Dark spot.

Thus, the correct option is a) one observes a dark spot.

Answer:

The answer is: A) a darkspot

Explanation:

When the red light passes through the glass plate, it is scattered. the convex lens (convergent lens) has two refractive surfaces, therefore, in the center of the lens, a characteristic dark spot would be observed.

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:

5N

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

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 current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

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Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

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In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.

Answers

When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

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