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Papessa [141]
2 years ago
11

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 9.8 s later its speed was 20 m/s in the oppo

site direction. What was the average acceleration of the particle during this 9.8 s interval
Physics
1 answer:
dem82 [27]2 years ago
6 0

The average acceleration of the particle in the time interval of 9.8 s is 10.204 m/s² opposite to the direction of motion.

Given:

Speed of particle, v₁ = 80 m/s (in positive x-direction)

Speed of particle, v₂ = -20 m/s (in opposite direction)

Time interval, Δt = 9.8 s

Calculation:

We know that, the average acceleration is given as:

a_avg = (v₂ - v₁)/ Δt     - ( 1 )

Applying values in above equation we get:

a_avg = (v₂ - v₁)/ Δt

           = (-20 m/s - 80 m/s) / (9.8 s)

           = -10.204 m/s²

Therefore the average acceleration of the particle in the time interval Δt is 10.204 m/s² opposite to the direction of motion.

Learn more about average acceleration here:

<u>brainly.com/question/27184422</u>

#SPJ4

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During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing
hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

d=v\times t

d=340\ m/s\times 5\ s  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

8 0
3 years ago
How many kcalories are provided by a food that contains 25 g carbohydrate, 6 g protein, and 5 g fat?
MrRissso [65]

<u>169 Kcalories</u> are provided by a portion of food that has 25 grams of carbs, 6 grams of protein, and 5 grams of fat.

Kcalories mean kilo-calories. Basically, kilo-calorie or kcal refers to 1,000 calories. To get the Kcalories of food, you have to add the kcal of carbohydrates, protein, and fat.

Get the product by multiplying the number of grams of carbohydrate, protein, and fat by 4,4, and 9, respectively. So if you want to get the energy or Kcal available from a meal, you must then combine the outcomes.

Simply put it, take note of the following conversions:

  • 1 gram of carbohydrate is 4kcal
  • 1 gram of protein is also 4kcal
  • Though, 1 gram of fat is 9kcal

So here's how to compute the Kcalories of food that contains 25g carbs, 6g protein, and 5g fat.

1.    25g x 4kcal/g = 100kcal

2.    6g x 4kcal/g = 24kcal

3.    5g x 9kcal/g = 45kcal

4.    100kcal + 24kcal + 45kcal = 169kcal!


Therefore, the food contains 169 kilo-calories!

You might be interested in nutrient density of an orange juice per kcalorie. Look here: brainly.com/question/26495283

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7 0
2 years ago
What is the name of the particle that compounds are made of?
VashaNatasha [74]

Answer:

Explanation:

atoms

8 0
3 years ago
Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea
notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, t=0.50\ s

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, v=0\ m/s

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, a=g=-9.8\ m/s^2

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

v=u+at

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0
3 years ago
Read 2 more answers
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
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