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Savatey [412]
3 years ago
8

A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning

(B) Negative Reinforcement (C) Spontaneous recovery (D) Positive Reinforcement
Physics
1 answer:
TEA [102]3 years ago
8 0
D. I think, not sure
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A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c
UkoKoshka [18]

Answer:0.00125 watts

Explanation:

resistance=50 ohms

Current=5 milliamps

Current=5/1000 milliamps

Current =0.005 amps

power=(current)^2 x (resistance)

Power=(0.005)^2 x 50

Power=0.005 x 0.005 x 50

Power=0.00125 watts

8 0
3 years ago
Read 2 more answers
A 93 kg zebra is traveling 13 m/s east. What is the zebra’s momentum?
Harlamova29_29 [7]
Equation: Mass x Velocity = Momentum

Answer: 93 x 13 = 1,209
3 0
3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force
bulgar [2K]

Te direction of the magnetic force for the velocity of the proton in the

-ve  y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule,  if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So,  we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x  direction,  so the magnetic force is acting in the positive z direction.

To know  more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

#SPJ4

4 0
2 years ago
A 4.44 l container holds 15.4 g of oxygen at 22.55°c. what is the pressure?
padilas [110]
We will use the ideal gas equation:
PV = nRT, where n is moles and equal to mass / Mr
P = mRT/MrV
P = 15.4 x 8.314 x (22.55 + 273) / 32 x 4.44
P = 266.3 kPa
5 0
3 years ago
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