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Leokris [45]
2 years ago
6

A soccer ball is released from rest at the top of a grassy incline. After 4.1 seconds, the ball travels 43 meters and 1.0 s afte

r this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant? Express your answer using two significant figures.
Physics
1 answer:
gavmur [86]2 years ago
5 0

Answer:

The acceleration of the ball as it moves down the grassy incline, if constant, is 5.1 m/s²

Explanation:

Initial velocity, u = 0m/s

Acceleration = ?

Vertical distance covered as at t=4.1s, H = 43m

Using the equations of motion,

H = ut + 0.5at²

43 = 0 + 0.5a (4.1)²

a = 43/(0.5×4.1²) = 5.1 m/s²

Hope this Helps!!

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You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios a
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Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 m/s^2

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion s=ut +\frac{1}{2}gt^2, we have

4.14 = 7.4t + 0.5\times9.8t^2

4.9t^2 + 7.4t - 4.14 =0

Using the quadratic formula \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

t = 0.43 s

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3 years ago
For a sound coming from a point source, the amplitude of the sound is inversely
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2 years ago
Mountain Ecosystems Assignment Answer each question below in complete sentences.1) Identify a predator from an Ethiopian Highlan
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An example of a a predator in the Ethiopian Highlands ecosystem is the

Wolf.The wolf is a carnivore which acts as a predator to other smaller

animals in the ecosystem.

<h3>What is a Prey?</h3>

Preys are mostly smaller animals in which the predators feed on for food.

They are usually herbivores and primary consumers in the ecosystem.

Examples of Preys include:

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Read more about Feeding relationship here brainly.com/question/9852437

8 0
1 year ago
A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi
sleet_krkn [62]

Answer:

Equilibrium temperature will be T=52.2684^{\circ}C

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So 2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)

0.334T-3.67=1688-32.031T

T=52.2684^{\circ}C

5 0
2 years ago
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