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3 years ago

True or False The greater the speed of an object, the less kinetic energy it possesses.

Physics

2 answers:

oksano4ka [1.4K]3 years ago

7 0

That Is A True Statement

Anettt [7]3 years ago

3 0

That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy

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Is position a base or derived quantity?

amid [387]

Position is measured in meters (m), so it is a base quantity.

<h3>What is base quantity?</h3>

A base or fundamental quantity is a physical quantity, in which other quantities are derived from.

Example of fundamental quantities;

Mass
Length (position)
Time
Temperature
Amount of substance

<h3>What is a derived quantity?</h3>

Derived quantities are those quantities obtained or expressed from fundamental quantities.

Example of derived quantities;

Speed
Acceleration
Volume
Area
Density, etc

Thus, we can conclude that position measured in meters (m) is a base quantity.

Learn more about base quantities here: brainly.com/question/14480063

#SPJ1

8 0

2 years ago

PLEASE HELP! It’s urgent... and please show your work!!

bagirrra123 [75]

Answer:

A) 3.8 x 10

Explanation:

4 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

yaroslaw [1]

Answer:

K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

p₀ = 0

final moment

p_f = m_a v_a + m_b v_b

p₀ = p_f

0 = m_a v_a + m_b v_b

tells us that

m_a = 8 m_b

0 = 8 m_b v_a + m_b v_b

v_b = - 8 v_a (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

Em₀ = K₀ = 73 J

final point. After separating the body

Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

K₀ = K_f

73 = ½ m_a (v_a² + v_b² / 8)

we substitute equation 1

73 = ½ m_a (v_a² + 8² v_a² / 8)

73 = ½ m_a (9 v_a²)

73/9 = ½ m_a (v_a²) = K_a

K_a = 8,111 J

3 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$