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3 years ago

Bandura's Bobo Doll Experiment concludes that _____.

1 answer:

3 0

The Bobo Doll Experiment was actually a collection of experiments that was done by Albert Bandura. These experiments involve observing the behavior of children upon seeing how an adult acts aggressively towards a Bobo Doll. Through this, it is concluded that children can indeed learn from the behavior done by adults. Based on this experimentation, it was concluded that boys are twice as likely as girls to act aggressively when seeing adults model aggression. The answer would be the first option.

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Which is a characteristic of projectile motion?

SVETLANKA909090 [29]

<h2>Right answer: It follows a curved path </h2>

The movement of a projectile is a movement in two dimensions (forming a curved path: a parabola shape) with constant acceleration.

A projectile is any body or object that is thrown or projected by means of some force and continues in motion by its own inertia. This means the only force that acts on it while in motion is the acceleration of gravity (in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$ ).

Where gravity influences the vertical movement of the projectile, while the horizontal movement of the projectile is the result of the tendency of any object to remain in motion at a constant speed (according to Newton's 1st law of motion sometimes called Law of Inertia).

The other options are incorrect because are false:

-The forward motion negates air resistance: There is always at least a small percent of air resistance, as long as that movement is done on Earth.

-It has variable acceleration: In projectile motion acceleration is constant (gravity acceleration) .

-It is unaffected by gravity: The only force that acts on the projectile is due gravity.

5 0

3 years ago

Read 2 more answers

What is the period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds

olga2289 [7]

The correct answer for this question is this one: C) 2.5s. The period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds is that 2.5 s

Here are the following choices:
A) 0.25s
B) 0.40s
C) 2.5s
D) 4.0s

7 0

3 years ago

A water skier lets go of the tow rope upon leaving the end ofa

ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

$v_v^2 + v_h^2 = v^2$

$v_v^2 + 13^2 = 14^2$

$v_v^2 = 196 - 169 = 27$

$v_v = \sqrt{27} = 5.2 m/s$

When the skier jumps to the its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv_v^2/2$

where m is the skier mass and h is the vertical distance traveled, $v_v$ is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

$gh = v_v^2/2$

$h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m$

3 0

3 years ago

A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu

romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

Condition (a);

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

Codition (b);

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0

3 years ago

For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz.

Natasha_Volkova [10]

Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance $C = 25.0\ \mu C$

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

$\omega = 2\pi f$

Put the value into the formula

$\omega =2\times3.14\times60.0$

$\omega=376.8\ rad/s$

(a). We need to calculate the value of $X_{L}$

Using formula of $X_{L}$

$X_{L}=\omega\times L$

Put the value into the formula

$X_{L}=376.8\times55.0\times10^{-3}$

$X_{L}=20.724\ \Omega$

(b). We need to calculate the value of $X_{L}$

Using formula of $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

$X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}$

$X_{C}=106.16\ \Omega$

(c). We need to calculate the value of Z

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

Put the value into the formula

$Z=\sqrt{75.0^2+(20.724-106.16)^2}$

$Z=113.68\ \Omega$

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{12.0}{75.0}$

$I=0.16\ A$

Using formula of rms current

$I_{rms}=\dfrac{I_{0}}{\sqrt{2}}$

$I_{rms}=\dfrac{0.16}{\sqrt{2}}$

$I_{rms}=0.113\ A$

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

$V_{rms}=0.113\times75.0$

$V_{rms}=8.475\ V$

(f). We need to calculate the rms voltage across the inductor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{L}$

$V_{rms}=0.113\times20.724$

$V_{rms}=2.342\ V$

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.113\times106.16$

$V_{rms}=11.99\ V$

(h). We need to calculate the dissipated power by the circuit

Using formula of dissipated power

$P=RI^2$

Put the value into the formula

$P=75.0\times0.113^2$

$P=0.958\ W$

Hence, This is the required solution.

3 0

3 years ago