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3 years ago

Which will result in a greater increase in kinetic energy, doubling the velocity or doubling the mass.

1 answer:

6 0

Kinetic energy is the energy possessed by a body while in motion. It is calculated by 1/2mv², where m is the mass of the body and v is the velocity.
Therefore, kinetic energy is dependent on both mass of the body and the velocity. An increase in mass increases the kinetic energy, an increase in velocity also increases kinetic energy of the body. Thus, doubling the mass and doubling the velocity will both increase the kinetic energy of the body.

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What units is the Metric System of Measurement based on?

Alex787 [66]

Answer:

metre for length and the kilogram for Mass

3 0

2 years ago

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

egoroff_w [7]

Answer:

4. both blocks will both have the same amount of kinetic energy.

Explanation:

When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.

Spring force is given as:

$F=k.\Delta x$

where: $\Delta x=$ length of compression in the spring

<u>We know kinetic energy is given by:</u>

$KE=\frac{1}{2} m.v^2$

Hence the kinetic energy of both the blocks is equal when they are released to move free.

8 0

3 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

3 years ago

A metal can become magnetic if the outermost electrons

Marizza181 [45]

Answer:

The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic behavior including paramagnetism, diamagnetism, and ferromagnetism.

An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number $m_s$ as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic.Explanation:

4 0

3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-

sergey [27]

$q = -21 * 10^{-6} C$

<h3>What is Free-fall acceleration?</h3>

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

$1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\$

and its sign must be negative so that it will have upward electric force

so it is

$q = -21 * 10^{-6} C$

The charge of a particle of mass is $-21 * 10^{-6} C$

Learn more about Free-fall acceleration: brainly.com/question/2165771

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8 0

2 years ago