A student's pencil rolls off a desk and lands 0.47 m away from the edge of the

During __________, frequently used synaptic connections in the brain strengthen while rarely used connections tend to thin.

Oxana [17]

The answer should be D. Synaptogenesis

8 0

3 years ago

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If a planet has an average distance from the sun (semi-major axis of its orbit) of 4 astronomical units, what is the period of i

alex41 [277]

Q. 380 If a planet has an average distance from the sun (semi-major axis of its orbit) of 4 astronomical units, what is the period of its orbit?

A. 1 years

B. 4 years

C. 64 years

D. 8 years

Q. 382 If a planet has an average distance from the sun (semi-major axis of its orbit) of 4 astronomical units, what is the period of its orbit?

A. 8 years

B. 1 years

C. 12 years

D. 64 years

Chapter 4; Problems 3, 5 3) If a planet has an average distance from the Sun of 4 AU, what is its orbital period? P2 = 43 = 64 P = 64 = 8 years

In conclusion:

The answer is 8 years

4 0

3 years ago

What type of machine are wire cutter pliers?

love history [14]

It a compound machine

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3 years ago

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A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s

aalyn [17]

Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a = 5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

7 0

3 years ago

An alpha particle (charge +2e) travels in a circular path of radius .5m in a magnetic field of 1.0 T. Find the (a) period, (b) s

navik [9.2K]

Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

Kinetic energy = 12.11 MeV

Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

qvB = m(v²/r)

qB = mv/r

v = qBr/m eq. 2

substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

r cancels out

T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

T = 0.13x10⁻⁶ seconds

(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

(c) kinetic energy (in electron volts)

Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV

5 0

3 years ago