Vary the sled’s height and mass. Observe the effect of each change on the potential energy of the sled.

a. How does potential energy change when height is increased?
b. How does potential energy change when mass is increased?
c. Compare a sled’s potential energy at 10 m to its potential energy at 20 m. How does doubling height affect potential energy?
d. Compare the potential energy of a 100-kg sled and a 200-kg sled at the same height. How does doubling mass affect potential energy?

Answer:

981N

Explanation:

The gravitational force between the football player and the Earth at the given distance is 979.91 N.

The given parameters:

The gravitational force between the football player and the Earth at the given distance is calculated by applying Newton's law of universal gravitation as follows.

[tex]F = \frac{GmM}{r^2} \\\\F = \frac{6.67 \times 10^{-11} \times 100 \times 5.98 \times 10^{24}}{(6.38 \times ^6)^2} \\\\F = 979.91 \ N[/tex]

Thus, the gravitational force between the football player and the Earth at the given distance is 979.91 N.

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Explanation:

A force involves an interaction between two or more objects, and it causes a push or pull between the objects. An example of opposing force include drag due to interaction with an air mass and the force due to friction between two objects.

hope it helps!

Answer:

D

Explanation:

Valley Formations

Answer:

C. Gas

Explanation:

Segment CD represents gaseous phase of matter as segment CD shows highest energy which is in case of gas.

Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.

Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.

The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.The intermolecular forces of attraction are different in different states of matter.Particles of matter have different sizes.

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#SPJ2

Answer:

the change in temperature..............

Answer:

1.5* 10¹⁵ Hz

Explanation:

        [tex]c = \lambda * f (1)[/tex]

       [tex]f = \frac{c}{\lambda} = \frac{3e8m/s}{200e-9m} = 1.5e15 1/sec = 1.5e15 Hz (2)[/tex]

Answer:

2√5N will be the weight.

Explanation:

Answer:

68 kg/m

Explanation:

bro

Answer:

90J

Explanation:

The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.

Answer:

D

Explanation:

a neutron does not have a positive nor negative charge it remains neutral

Answer:

-4.35 × 10^-6 N

Explanation:

i just answered it on ap3x :)

Answer:

the kinetic energy of body B is twice the kinetic energy of body A

Explanation:

The kinetic energy of a body is given by

          K = ½ m v²

If two objects leave the same point, suppose that at the same height when they reach the ground they have the same velocity.

Therefore if the mass of body b is twice the mass of body A

          [tex]m_{b} = 2 m_{a}[/tex]

         [tex]K_{b}[/tex] = ½ (2 [tex]m_{a}[/tex]) v²

          K_{b} = 2 (½ m_{a} v²)

          K_{b} = 2 K_{a}

therefore the kinetic energy of body B is twice the kinetic energy of body A

Answer:

1962 joules

Explanation:

m = 10 kg

h = 20 m

g = 9.81 ms^-2

PE = ?

PE = MGH

PE = 10 x 9.81 x 20

PE = 1962 joules

Answer:

[tex]545.89\ \text{Hz}[/tex]

Explanation:

[tex]v_o[/tex] = Velocity of observer = 0

[tex]v_s[/tex] = Velocity of source = 34 m/s

v = Speed of sound in air = 343 m/s

f = Emitted frequency = 600 Hz

Doppler effect will be observed

[tex]f_o=f(\dfrac{v+v_o}{v+v_s})\\\Rightarrow f_o=600(\dfrac{343+0}{343+34})\\\Rightarrow f_o=545.89\ \text{Hz}[/tex]

The frequency heard will be [tex]545.89\ \text{Hz}[/tex].

Answer:

Magnitude of the force = 1.25 N

Explanation:

Given:

Spring constant k = 50.0 N/m

Distance s = 2.50 cm = 0.025 m

Find:

Magnitude of the force

Computation:

Magnitude of the force = Ks

Magnitude of the force = 50 x 0.025

Magnitude of the force = 1.25 N

Answer:

acceleration=1.6m/s²

Explanation:

F=m×g

g=F/m

g=80N/50=1.6m/s²

Answer:

No.

Explanation:

Acceleration of an object is defined as the rate of change of velocity per unit time. It can be given by :

a = dv/dt

dv is the change in velocity i.e. final velocity-initial velocity

dt is change in time

At a particular time, its change in velocity can be one only. It would imply that a body cannot have two acceleration at a particular time.

Answer:

[tex]149.112\ \text{N}[/tex]

[tex]120.94\ \text{N}[/tex]

[tex]302.77\ \text{N}[/tex]

[tex]58.44\ \text{N}[/tex]

Explanation:

m = Mass of block = 15.2 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]\theta[/tex] = Angle

a = Acceleration

F = Applied force

Normal force is given by

[tex]N=mg\\\Rightarrow N=15.2\times 9.81\\\Rightarrow N=\boldsymbol{149.112\ \mathbf{N}}[/tex]

[tex]\theta=35^{\circ}[/tex]

[tex]N=mg\cos\theta\\\Rightarrow N=15.2\times 9.81\times \cos35.8^{\circ}\\\Rightarrow N=\mathbf{120.94\ N}[/tex]

[tex]a=3.53\ \text{m/s}^2[/tex]

[tex]N=m(g+a)\\\Rightarrow N=15.2(9.81+3.53)\\\Rightarrow N=\boldsymbol{302.77\ \mathbf{N}}[/tex]

[tex]F=155\ \text{N}[/tex]

[tex]N=mg-F\sin35.8^{\circ}\\\Rightarrow N=15.2\times 9.81-155\sin35.8^{\circ}\\\Rightarrow N=\boldsymbol{58.44\ \mathbf{N}}[/tex]

Answer: Option C.

Explanation:

When the box is moving, he pushes with a force F.

in this case, the coefficient is μk, and the box is accelerating, this means that the net force is not zero, so F is larger than the friction force.

When the box is still, the man pushes again with a force F, but now the box does not move, so there is no acceleration, which means that the net force is zero, then F is not greater than the maximum static friction force.

Now, the friction force between an object of mass M, and a surface with a coefficient of friction  μ is:  μ*m*g

where g is the gravitational acceleration.

Then, from the first part, we can conclude that:

μk*m*g < F

(the force F is larger than the kinetic friction force)

and from the second part, we know that:

F ≤  μs*m*g

(The force F is not greater than the static friction force)

If we write those two together, we have:

μk*m*g < F ≤  μs*m*g

Then the correct option is c.

Answer: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.

Explanation:

That's all of the planets if you need them. Hope this helps!

Answer:

30 mm

Explanation:

[tex]\lambda[/tex] = Wavelength of light = 503 nm

d = Diameter of hole = 0.2 mm

D = Distance of screen from light source = 4.9 m

Diameter of the central bright area is given by

[tex]x=\dfrac{2.44\lambda D}{d}\\\Rightarrow x=\dfrac{2.44\times 503\times 10^{-9}\times 4.9}{0.2\times 10^{-3}}\\\Rightarrow x=0.03\ \text{m}=30\ \text{mm}[/tex]

The diameter of the central bright area is 30 mm.

It is given that up direction be positive.

Ball velocity at the time is was dropped is 3 m/s.

g = -9.8 m/s²

s = -50 m

Now, using equation of motion for time :

s = ut + at²/2

3t - 4.9t² + 50 = 0

4.9t² - 3t - 50 = 0

Solving above equation, we get :

t = 3.52 s

Therefore, the rock will splash into the water after 3.52 s.

Answer:

Median

Explanation:

Extra information:-

Median formula:-

[tex]\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}[/tex]

Answer:

The answer is above this

Explanation:

Answer:

a) The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle at point B is approximately 5.270 meters per second.

c) The total work done on the particle as it moves from A to B is - 9.78 joules.

Explanation:

Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.

a) The translational kinetic energy of the particle ([tex]K[/tex]), measured in joules, is determined by the following formula:

[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Speed, measured in meters per second.

If we know that [tex]m = 0.54\,kg[/tex] and [tex]v = 8\,\frac{m}{s}[/tex], the translational kinetic energy at point A is:

[tex]K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}[/tex]

[tex]K = 17.28\,J[/tex]

The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle is clear in (1):

[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]

If we know that [tex]K = 7.5\,J[/tex] and [tex]m = 0.54\,kg[/tex], then the speed of the particle at point B:

[tex]v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }[/tex]

[tex]v\approx 5.270\,\frac{m}{s}[/tex]

The speed of the particle at point B is approximately 5.270 meters per second.

c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ([tex]W_{A\rightarrow B}[/tex]), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:

[tex]W_{A\rightarrow B} = K_{B}-K_{A}[/tex] (2)

If we know that [tex]K_{A} = 17.28\,J[/tex] and [tex]K_{B} = 7.50\,J[/tex], then the change in the translational kinetic energy of the particle is:

[tex]W_{A\rightarrow B} = 7.50\,J-17.28\,J[/tex]

[tex]W_{A\rightarrow B} = -9.78\,J[/tex]

The total work done on the particle as it moves from A to B is - 9.78 joules.

Answer:

The correct answer is a

Explanation:

The electric field is given by the relation

         F = q E

where The force is the Coulomb force and q is a positive test charge, therefore the electric field has the same direction as the electric force.

Consequently if the charge is positive the field must go out of the charge, if the charge is negative the electric field must be directed towards the charge.

Consequently, in this exercise we are told that the lines are directed towards the load, therefore the load must have a negative sign.

The correct answer is a

Answer:

A) The disk arrives first

Explanation:

The disk arrives first, and the reason behind this is that the disk posses smaller moment of inertia, picking up quickly of angular speed of an object for a particular torgue can be determined by ts moment of inertia . So small quantity of energy will be used up as rotational energy, and with the rotational energy the disk will move faster that the hoop.

The option that illustrates the object that reaches the bottom first is A. The disk arrives first.

From the information given, we're told that a solid disk and a hoop are simultaneously released from rest at the top of an incline and rolled down without slipping.

It should be noted that the disk will arrive first due to the fact that it has smaller inertia. In such a case, the disk will arrive first while the hoop arrives later.

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Given :

Force provided, F = 112 N.

Acceleration of the bag, a = 0.79 m/s².

To Find :

a. What is the mass of the fertilizer bag?

b. How much does the fertilizer bag weigh?

Solution :

We know, force is given by :

F = ma

m = F/a

m = 112/0.79 kg

m = 141.77 kg

Now, weight is given by :

W = mg

W = 141.77 × 9.8 N

W = 1389.35 N

Therefore, the mass of fertilizer bag is 141.77 kg and weight us  1389.35 N.

Total distance = 36500 m

The average velocity = 19.73 m/s

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m[/tex]

[tex]\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s[/tex]

State 2 : constant speed

[tex]\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m[/tex]

State 3 : deceleration

[tex]\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)[/tex]

[tex]\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m[/tex]

Total distance : state 1+ state 2+state 3

[tex]\tt 200 + 36000 + 300=36500~m[/tex]

the average velocity = total distance : total time

[tex]\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s[/tex]

The total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.

What is average velocity?

The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.

The given data in the problem is;

v₀ is the initial velocity=0(from rest)

a is the acceleration= 1 m/s²

t₁ is the time period 1= 20 s

t₂ is the time period 2= 0.5 hr = 1800 s

t₃ is the time period 3= 30 s

For part 1

The distance travelled is found as;

[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=\frac{1}{2}\times 1 \times (20)^2 \\\\ \rm d=200\ m[/tex]

The initial velocity is found as ;

[tex]\rm v_t=v_0+at\\\\ \rm v_t=0+1 \times 20\\\\ \rm v_t=20 \ m/sec[/tex]

For part 2

The distance traveled is found as;

[tex]\rm d= v \times t \\\\ \rm d= 20 \times 1800 \\\\ \rm d= 36000 \ m[/tex]

For part 3

The acceleration is found as;

[tex]\rm v_t=v_0+at\\\\ \rm 0=v_0+a \times 30\\\\ \rm 20=-30 a \\\\ a= - \frac{2}{3}\ m/sec^2[/tex]

The distance traveled is found as;

[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=20.30 +\frac{1}{2} \times \frac{2}{3} (30)^2 \\\\ \rm d=300 \ m[/tex]

Total distance travelled= distance travelled( part 1+ part 2+ part3)

Total distance travelled=200+3600+300 = 36500 m

The average velocity is found as;

[tex]v_{avg}= \frac{36500}{20+1800+30} \\\\ v_{avg}= 19.73 \ sec[/tex]

Hence the total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.

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Answer:

500 N

Explanation: