A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.

At what point is its potential energy greatest?

At what points does it have zero kinetic energy?

At what point did it have maximum kinetic energy?

Answers

Answer 1

Answer:

Greatest potential: moment before being dropped

Zero Kinetic: when it comes to rest

Greatest Kinetic: moment before first bounce

Explanation:


Related Questions

if Petrol diesel etc catches fire one should never try to extinguish in using water why?​

Answers

Answer:

because both petrol and diesel are oil

Explanation:

oil floats on water that's why if we will try to extinguish with water so the fire will float on water

hope u like my answer

please mark methe brainest

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.

What is the magnitude of the velocity after it hits the ground?

Answers

Answer:

9.25 m/s

Explanation:

A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball

Answers

Answer:

The maximum height risen by the bullet-baseball system after the collision is 81.76 m.

Explanation:

Given;

mass of the bullet, m₁ = 0.033 kg

mass of the baseball, m₂ = 0.15 kg

initial velocity of the bullet, u₁ = 222 m/s

initial velocity of the baseball, u₂ = 0

let the common final velocity of the system after collision = v

Apply the principle of conservation of linear momentum to determine the common final velocity.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

0.033 x 222   +  0.15 x 0     = v(0.033 + 0.15)

7.326  =  v(0.183)

v = 7.326 / 0.183

v = 40.03 m/s

Let the height risen by the system after collision = h

Initial velocity of the system after collision = Vi = 40.03 m/s

At maximum height, the final velocity, Vf = 0

acceleration due to gravity for upward motion, g = -9.8 m/s²

[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]

Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.

The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Answers

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

convert 2.4 milimetre into metre​

Answers

Answer is 0.0024

Explanation

divide the length value by 1000.

what is the force of a body which have mass of 7 kg​

Answers

Answer:

Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N

Answer:

10 m/s2 or 70 newtons.

Explanation:

............................

............

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.

Answers

Answer:

sup qwertyasdfghjk

Explanation:

An electron in a hydrogen atom is in a p state. Which of the following statements is true?


a.
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).



b.
The electron has an energy of -13.6 eV.


c.
The electron has a total angular momentum of ħ.


d.
The electron has a z-component of angular momentum equal to sqrt(2)* ħ.

Answers

Answer:

The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).

Explanation:

We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.

P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.

Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node

Help me with my physics, please

Answers

The right answer would be

-20t+ 80

You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450 m with a radial acceleration of 17 m/s^2.

Required:
What is the plane's speed?

Answers

Answer:

v = 87.46 m/s

Explanation:

The radial acceleration is the centripetal acceleration, whose formula is given as:

[tex]a_c = \frac{v^2}{r}[/tex]

where,

[tex]a_c[/tex] = centripetal acceleration = 17 m/s²

v = planes's speed = ?

r = radius of path = 450 m

Therefore,

[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]

v = 87.46 m/s

A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?

Answers

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the
acceleration of the block?

Answers

Answer:

F = m g sin theta      force accelerating block

m a = m g sin theta

a = 9.8 sin 24 = 3.99 m/sec^2

Increased air pressure on the surface of hot water tends to
A) prevent boiling.
B) promote boiling.
C) neither of these

Answers

Answer

The answer is B

Explanation

It is B because If the more pressure of the earths surface it is promoting more pressure and it is boiling so it’s B

Hope this helps :)

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info

Answers

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

Find the volume of cuboid of side 4cm. Convert it in SI form​

Answers

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

I need help with this problem can anybody help me please , it’s physics 2 course

Answers

Answer:

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Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to

Answers

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

x₁ = -0.20 mx₂ = 0 mk = 60 N/m

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

Answers

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

two resistors with resistance values 4.5 ohms and 2.3 ohms are connected in series or parallel across a potential difference of 30V to a light bulb find the current flowing through the light bulb in both cases​

Answers

Answer:

Look at work

Explanation:

Series:

I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.

Parallel:

V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.

This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.

Two bar magnet with the north pole of one near the south pole of the second. field lines are leaving the north pole and bent away from the south pole of the other. Field lines are leaving the south pole of one and bending away from the north pole of the other.
Which change will correct the error in the diagram?

a)changing the N to S
b)reversing the arrows on the left to point toward the N
c)changing the S to N
d)reversing the arrows on the right to point toward the S

Answers

Answer:

changing the N to S. that's how the error will be corrected

Answer:

C is the correct answer

Explanation:

i took the test

crushing chalk into powder is and irreversible change. is this example a physical or chemical change?Why?​

Answers

Answer:

It is a example of physical change

Two objects are interacting, but stay stationary. Which best describes what is happening to the action and react
forces?
There are no forces acting on the objects.
The forces are equal and opposite each other.
One object is exerting more force than the other.
O Their forces are occurring at different times.

Answers

Awnser is B. The forces are equal opposite of each other.

Answer:b the force are equal and opposite each other

Explanation:

When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)

Answers

P = F v

where P is power, F is the magnitude of force, and v is speed. So

50.0 hp = 37,280 W = F (24.6 m/s)

==>   F = (37,280 W) / (24.6 m/s) ≈ 1520 N

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request

Answers

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

What best describes a societal law

Answers

Answer:

Societal laws are based on the behavior and conduct made by society or government.

hope it helps.stay safe healthy and happy.

In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?

Answers

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.

Answers

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.

how do you calculate voltage drop

Answers

Answer:

Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.

Divide by 100.

Multiply by proper voltage drop value in tables. The result is voltage drop.

Explanation:

d. On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg>m3 . If the tank was full before the accident, what was the total outward force the molasses exerted on its sides

Answers

Answer:

F = 1.638 x 10⁸ N = 163.8 MN

Explanation:

The total force exerted by the molasses is given as:

F = PA

where,

F = Force exerted by the molasses = ?

P = Pressure = ρgh

ρ = density of molasses = 1600 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h = height of tank = 17.7 m

A = cross-sectional area of tank = πr²

r = radius of tank = 27.4 m/2 = 13.7 m

Therefore,

[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]

F = 1.638 x 10⁸ N = 163.8 MN


Question 8 a-e plz

Answers

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

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