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Leokris [45]
3 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]3 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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Amadeo the Archer is practicing his archery skills. He shoots both arrows into a field. His first arrow (arrow A) is embedded ha
attashe74 [19]

Answer:

1

The arrow with greater impart is Arrow B

2

The both arrows will feel the same impulse

Explanation:

1. Arrow B since KE= W = F * d  it used more force to stop itself in a shorter distance.

2. They should feel the same impulse since the both had the same momentum    m * v= F * t

4 0
3 years ago
liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i
aksik [14]

4.80 \times 10^3 \text { seconds }  long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

#SPJ4

3 0
1 year ago
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735
Irina18 [472]
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v =  speed of pumping the gasoline, m/s
Then the mass flow rate is 
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer:  2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
7 0
3 years ago
Read 2 more answers
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

      520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2

Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
A book is sitting on a table. Which of the following is true about the table? O A. It is pushing up on the book. O B. It exerts
777dan777 [17]

Answer:

yes it does exert a force, it pushes it up

Explanation:

this is called normal force

if it didn't exert a force the book would keep going down

according to newton every force has an equal amd opposite reaction

so the book exerts a force on the table and vice versa

hope this helped

5 0
3 years ago
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