Please see the Newton's 2nd Law which states that an object accelerates if there is a net or unbalanced force on it. In this scenario there is just one force exerted on the wagon i.e: the force that the horse exerts on it. The wagon accelerates because the horse pulls on it. And the amount of acceleration equals the net force on the wagon divided by its mass.

As there are two forces the push and pull the horse; the wagon pulls the horse backwards, and the ground pushes the horse forward. The net force is determined by the relative sizes of these two forces.

If the ground pushes harder on the horse than the wagon pulls, there is a net force in the forward direction, and the horse accelerates forward, and if the wagon pulls harder on the horse than the ground pushes, there is a net force in the backward direction, and the horse accelerates backward.

If the force that the wagon exerts on the horse is the same size as the force that the ground exerts, the net force on the horse is zero, and the horse does not accelerate.

7 0

3 years ago

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Write an expression to find how many times Jason will blink in 60 minutes if he continues to blink 19 times a minute.

Travka [436]

Y= 19x & x is 60 so you just do 19 times 60n

3 0

3 years ago

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A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that a

madam [21]

Answer:

Option D. 938/969

Step-by-step explanation:

At most 2 defective means 2 or less than 2 bulbs are defective

So, we have 3 cases:

a) No defective bulb. b) 1 defective bulb. c) 2 defective bulbs

Case a) No defective bulb

Total number of bulbs = 20

Number of defective bulbs = 5

Number of non-defective bulbs = 15

Total number of ways to select 0 defective bulb = 15C4 = 1365

Case b) 1 defective bulb

Total number of ways to select 1 defective bulb = 15C3 x 5C1 = 2275

Case c) 2 defective bulbs

Total number of ways to select 2 defective bulbs = 15C2 x 5C2 = 1050

Therefore, total number of ways to select at most 2 defective bulbs = 1365 + 2275 + 1050 = 4690

Total number of ways to select 4 bulbs from 20 = 20C4 = 4845

Therefore, probability of selecting at most 2 defective bulbs = $\frac{4690}{4845}=\frac{938}{969}$

Therefore, option D gives the correct answer.

6 0

3 years ago