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3 years ago

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Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant condit

ioning B. determinism C. self-efficacy D. self-worth

2 answers:

Darya [45]3 years ago

4 0

Answer hopefully = C. self-efficacy

Rzqust [24]3 years ago

4 0

Answer:

C

Explanation:

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The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug

Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

<u>Energy of Photon = 4.091 MeV</u>

8 0

3 years ago

A trained eye in the dark for an extended period of time may pick up a light stimulus from a light source, at the lowest radiate

frez [133]

Answer:

#_photons = 30 photons / s

Explanation:

Let's start by finding the energy of a photon of light, let's use the Planck relation

E = h f

the speed of light is related to wavelength and frequency

c = λ f

we substitute

E = h c /λ

E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹

E₀ = 3.978 10⁻¹⁹ J

now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second

#_photons = 1 photon (E / Eo)

#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹

#_photons = 3.0 10¹

#_photons = 30 photons / s

3 0

3 years ago

A ball is dropped from rest. how fast is the ball going after 3 seconds

Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

5 0

3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

$\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}$

$\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}$

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

= 257.23/2.0

= 128.61 Watts

8 0

3 years ago

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

Montano1993 [528]

Answer:

vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

W = ΔK (1)

for this part the mass is

M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

N -W = 0

N = W = Mg

friction forces have the expression

fr =μ N

fr = μ M g

we substitute in 1

-μ M g x = 0 - ½ M v²

v² = 2 μ g x

let's calculate

v² = 2 0.750 9.8 6.00

v = ra 88.5

v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

p_f = (mₐ + m_b) v

p₀ = p_f

+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

660 vₐ₀ = (660 +490) 9.39 + 490 17.778

vₐ₀ = 19509.72 / 660

vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0

3 years ago