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vova2212 [387]
2 years ago
13

a 2.5 kg rock is dropped off a 32 m cliff and hits a spring, compressing it 57cm. what is the spring constant

Physics
2 answers:
Rama09 [41]2 years ago
4 0

2.5 kg because you cant change the weight of the rock

Mila [183]2 years ago
3 0

The answer is 4,800N/m

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A catcher stops a 0.15-kg ball traveling at 40 m/s in a distance of 20 cm. what is the magnitude of the average force that the b
iren2701 [21]

The magnitude of the average force that the ball exerts against his glove is 600 N

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.15 kg

initial speed of ball = u = 40 m/s

final speed of ball = v = 0 m/s

distance = d = 20 cm = 0.2 m

<u>Asked:</u>

average force = F = ?

<u>Solution:</u>

<em>We will use </em><em>Newton's Law of Motion</em><em> to solve this problem as follows:</em>

F = m a

F = m (\frac { u^2 - v^2 } { 2d } )

F = 0.15 \times \frac { 40^2 - 0^2 } { 2 \times 0.2 }

F = 0.15 \times \frac { 1600 } { 0.4 }

F = 0.15 \times 4000

\boxed {F = 600 \texttt{ N}}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
Read 2 more answers
g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the
Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

R = u^2 / g

5 0
2 years ago
Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13
Julli [10]
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
                 m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
                    (85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
3 0
2 years ago
If you lived on Saturn, which planets would exhibit retrograde motion like that observed for Mars from Earth? (Select all that a
liberstina [14]

Answer:

a) Earth

b) Mercury

c) Neptune

Explanation:

All the planets move around the sun in eastward direction, but few planet have retrograde rotation i.e in westward direction. Retrograde motion is just an apparent change in the movement of planet which means it only seems as if the planet are rotating in opposite direction. Retrograde movement of planet like  Saturn, Jupiter and mars is not real. Hence, if a person lives on Saturn, then following planets will exhibit retrograde motion  

a) Earth

b) Mercury

c) Neptune

4 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
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