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kkurt [141]
3 years ago
13

Raven throw a baseball directly downward from a terrace froma speed of 5.0 m/s. How fast it will be moving when it hits the path

way 3.0 m below
Physics
1 answer:
fredd [130]3 years ago
7 0

Answer:

The speed of the ball at this distance is 9.15 m/s

Explanation:

Given;

initial speed of the baseball, U = 5.0 m/s.

distance traveled along the path way, h = 3 m

final speed of the baseball at this distance, V = ?

The baseball is falling under the influence of gravity.

Acceleration due to gravity, g is positive, since the baseball is falling towards its direction.

g = 9.8 m/s²

Apply the third kinematic equation;

V² = U² + 2gh

V² = 5² + 2 x 9.8 x 3

V² = 25 + 58.8

V² = 83.8

V = √83.8

V = 9.15 m/s

Therefore, the speed of the ball at this distance is 9.15 m/s

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A cheetah can run at approximately 100 km/hr and a gazelle at 80 km/hr. If both animals are running at full speed, with a gazell
zimovet [89]
The cheetah's speed is 100x and The gazelle's speed is 80x + 70. Set the two equations equal to each other: 100x = 80x +70 (then subtract 80x from both sides). 20x = 70 (then divide by 20). X =3.5. The cheetah catches the gazelle after 3.5
6 0
4 years ago
The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k
Murrr4er [49]

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2

Force is given by

F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N

The force with which the team pulls the plane is 28716.4740661 N

v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

Work done is given by

W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J

The fraction is given by

\dfrac{54450.9540448}{106250.954045}=0.512474965841

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

3 0
3 years ago
What is happening at the atomic level to give rise to the observed energy?
ziro4ka [17]
Here is the answer. What is happening at the atomic level to give rise to the observed energy is that t<span>he </span>atomic level<span> is affected by the movement of electrons so as to </span><span>give rise to the observed energy. Hope this answers your question. Have a great day!</span>
6 0
4 years ago
How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
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