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3 years ago

How do the !Kung Bushmen teach their children not to be violent?

2 answers:

7 0

Answer:Kung Bushmen teach their children not to be violent? Creating a culture that chooses non-violence with intention ... Kung case, parents are not likely to reach the point of abusing their children, but in the unlikely event that someone did .

Explanation:May i plz have brainlist only if u wanna give me brainlist though have an nice day!

makkiz [27]3 years ago

7 0

Answer:

By not reacting to tantrums like this, children of the !Kung tribe learn that adults will not react. The adults will not act differently toward the child who is acting out. Having a tantrum and acting aggressively will not get the adult’s attention or sympathy. The children learn that they will get no reward, such as attention, from their parents for acting out. They learn that aggression and fighting will not get them anywhere. Draper’s time with the !Kung also showed her that children who are raised in this way learn to be more cooperative rather than competitive. The children of the tribe grew up in mixed-age groups. By not being surrounded by many kids their own age, they felt less desire to compete. The older children were pleasant and calm when dealing with the younger kids. Children of all ages experienced working together as they grew up. They enjoyed accomplishing different tasks as a group.

Explanation:

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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1

Shkiper50 [21]

Answer:

0.10013 atm

Explanation:

Applying Boyle's Law,

P'V' = PV................... Equation 1

Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.

make P the subject of the equation

P = P'V'/V..................... Equation 2

Given: P' = 0.355 atm, V' 0.110 m³, V = 0.390 m³

Substitute into equation 2

P = 0.355(0.11)/0.39

P = 0.10013 atm.

5 0

3 years ago

A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

4 0

3 years ago

A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

T = F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm $= \frac{2}{100} = 0.02m$ for z the perpendicular distance

So Elbow Torque is $T_e= 4000 * 0.02$

$= 80Nm$

To obtain the torque required we substitute 300 N for F and 30cm $=\frac{30}{100} = 0.3 m$

So the Required Torque is $T_R = 300 *0.3$

$=90Nm$

Now since $T_e < T_R$ it mean that the weight lifter would not get past this sticking point

7 0

3 years ago

A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi

oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775°

The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25

F net = 49N

Work is product of net force and distance:
W = F net * d = 49 * 20

Work = 980 J

4 0

3 years ago