An 80-kg man balances the boy on a teeter-totter as shown. What is the approximate mass of the boy? What, approximately, is the magnitude of the downward force exerted on the fulcrum? Ignore the weight of the board.
I know the first answer is 20 kg and the second answer is 1000N. Have no idea how to get these answers though. Please help? Will rate!!

Answer:

Initially the PE of the object was W * h = 30 * 10 = 300 Joules

The KE of the object when it struck the ground was 1/2 M v^2

KE = 1/2 * 30/9.8 * 13^2 = 259 J

So the object lost 41 J to friction during the fall

Since Work = Force * distance

Force = 41 J / 10 m = 4.1 N  (the average force of friction)

Answer:

Explanation:

Above is the unscrambled words. Hope this helps!

Answer:

afap

Explanation:

Answer:

option ( a ) is correct .

Explanation:

Escape velocity on the earth = √ ( 2 GM / R )

where G is universal gravitational constant , M is mass of the earth and R is radius .

V₀ = √ ( 2 GM / R )

escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth

Radius of the planet = 4 R

escape velocity of planet = √ ( 2 GM / 4R )

= .5 x √ ( 2 GM / R )

= .5 V₀

option ( a ) is correct .

C) The mantle's particles will become more dense and rise as a result of the core heat, causing the lithosphere's plates to move in the opposite direction of the convection current.

Convection Currents The lithosphere becomes denser than the surrounding magma and sinks back toward the core as the heated mantle transfers heat energy to it.

Conduction from the core directly heats the lower mantle. As atoms collide, heat is transferred in conduction. Convection is the movement of warm and cool materials upward and downward. Material that is hot rises to the surface through conduction.

“hot spots," which are non-plate tectonic volcanic regions, are probably caused by mantle plumes. A diapir is formed when a mantle plume reaches the upper mantle. Volcanic eruptions are sparked by the heat generated by this molten material in the asthenosphere and lithosphere.

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The gravitational field strength on Oobla is 20.75 m/s².

The gravitational field is the gravitational force exerted per unit mass on a small mass at a point in the field.

To calculate the gravitational field strength on Ooobla, we use the formula below

Formula:

Where:

From the question,

Given;

Substitute these values into equation 1 and solve for g'

Hence, the gravitational field strength is 20.75 m/s².

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Answer:

a)[tex]W=0.8J[/tex]

b) [tex]d_t=0.20m[/tex]

c) [tex]\triangle K.E=0.267J[/tex]

Explanation:

From the question we are told that:

Mass of cart 1 [tex]M_1=1.0kg[/tex]

Mass of cart 1 [tex]M_2=0.05kg[/tex]

Force on  cart 1 [tex]F_1=4.0N[/tex]

Push Distance of cart 1 [tex]d_1=0.20m[/tex]

Slide Distance of cart 1 [tex]d_1'=0.35m[/tex]

a)

Generally the equation for work-done is mathematically given by

[tex]W=f*d\\W=4*0.20\\W=0.8J \\[/tex]

b)

The systems center of mass moved a net totally of (while being pushed)

Mass 1 =0.20m

Mass 2=0

Therefore

[tex]d_t=d_1+d_2[/tex]

[tex]d_t=0.20+0[/tex]

[tex]d_t=0.20m[/tex]

c)

Since work-done is equal to K.E energy of cart 1

Therefore

[tex]W=1/2mv^2[/tex]

[tex]V_1=\sqrt{\frac{W}{1/2m}}[/tex]

[tex]V_1=\sqrt{\frac{0.8}{1/2(1)}}[/tex]

[tex]V_1=1.264[/tex]

Therefore Kinetic energy before collision is

[tex]K.E_1=1/2mv^2[/tex]

[tex]K.E_1=1/2*1*1.264^2[/tex]

[tex]K.E_1=0.768[/tex]

Generally from the equation for conservation  of momentum the Velocity of cart 2 is mathematically given by

[tex]v_2=\frac{m_1V_1}{m_1+m_2}[/tex]

[tex]v_2=\frac{1*1.264}{1+0.5}[/tex]

[tex]V_2=0.842m/s[/tex]

Therefore the final K.E is mathematically given by

[tex]K.E_2=(1/2)(m_1+m_2)V_2^2[/tex]

[tex]K.E_2=1/2*(1.5)(0.842)^2[/tex]

[tex]K.E_2=0.531J[/tex]

Generally the Change in K.E is mathematically given by

[tex]\triangle K.E=K.E_1-K.E_2[/tex]

[tex]\triangle K.E=0.798-0.531[/tex]

[tex]\triangle K.E=0.267J[/tex]

Therefore the will force change the kinetic energy of the system's center of mass by

[tex]\triangle K.E=0.267J[/tex]

(a) The work done by you when you push the cart at a constant force is 0.8 J.

(b) The distance moved by the center mass of the two cart system is 0.23 m.

(c) The change in kinetic energy of the system center of mass is 0.271 J.

The work done by you when you push the cart at a constant force is calculated as follows;

W = Fd

W = 4 x 0.2

W = 0.8 J

Xcm = (m₁x₁ + m₂x₂)/(m₁ + m₂)

Xcm = (0.5(0) + 1(0.35)) / (1 + 0.5)

Xcm = (0.35)/(1.5)

Xcm = 0.23 m

F = ma

a = F/m

a = (4)/1 = 4 m/s²

v² = u² + 2as

v² = 0 + 2(4)(0.2)

v² = 1.6

v = √1.6

v = 1.265 m/s

m₁u₁ + m₂u₂ = v(m₁ + m₂)

1(1.265) + 0 = v(1 + 0.5)

1.265 = 1.5v

v = 0.84 m/s

K.E(initial) = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E(initial) = ¹/₂(1)(1.265)² + ¹/₂(0.5)(0) = 0.8 J

K.E(final) = ¹/₂(m₁ + m₂)v²

K.E(final) = ¹/₂(1 + 0.5)(0.84)²  = 0.529 J

Δ K.E = 0.8 J - 0.529 J = 0.271 J

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Answer:

a)   [tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)   [tex]V=1.8217*10^{-3}[/tex]

Explanation:

From the question we are told that:

Mass of woman [tex]M_w=55kg[/tex]

Radius of merry go round [tex]r=2.0m[/tex]

Rotational inertia [tex]i= 1250 kg-m2[/tex]

Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]

Speed of rock [tex]V_r=2.0m/s[/tex]

Tangent angle to the outer edge [tex]\theta=1[/tex]

a)

Generally the equation for conservation of momentum is mathematically given by

[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]

[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]

[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]

[tex]\omega=9.10*10^{-4}rad/sec[/tex]

b)

Generally the equation for linear speed V is mathematically given by

[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]

[tex]V=1.8217*10^{-3}[/tex]

Complete Question:

A horizontal force of 5N is required to maintain a velocity of 2m/s for a block of 10kg mass sliding over a rough horizontal surface. The work done by this force in one minute is:

Answer:

Workdone = 600 Joules.

Explanation:

Given the following data;

Force = 5N

Velocity = 2 m/s

Time = 1 minute to seconds = 60 seconds.

Mass = 10 kg

To find the work done;

First of all, we would determine the distance covered by the block.

Distance = speed * time

Distance = 2 * 60

Distance = 120 seconds

Mathematically, workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

Substituting into the formula, we have;

[tex] Workdone = 5 * 120 [/tex]

Workdone = 600 Joules.

Therefore, the work done by this force in one minute is 600 Joules.

The crate will be under a net force of 95 N. Resolving the force in the x-direction yields the net force on the crate: F = 300 cos 10° - N F= 295.44-196.45, F= 95.

The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value. The formula fr = Fr/N serves as a representation of it.

The friction of a moving item is inversely proportional to the normal force and runs perpendicular to it. The type of surface the thing comes into touch with determines the amount of friction it experiences. As long as there is a point of contact, friction does not depend on the area of contact.

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The resultant (x,y) coordinate of vector B is  (- 5, 3) and that of A is (9,3). Then, the sum of x components of vector A and B is 4.

Vectors are physical quantities having both magnitude and direction. For example force, velocity, displacement etc .are vector quantities. Coordinates are used to describe the position of these vectors.

The coordinates of tail of  the vector A is (0, 0) and the head is (9,3)

Then the coordinate of A is (9 -0, 3-0) = (9,3)

Similarly, the tail coordinate of B is (9,3) and head is (4,6). The coordinates of B vector is (4-9, 6-3) = (-5, 3)

Now the sum of x components of A and B is = - 5 + 9 = 4.

Therefore, option b is correct.

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Answer:

B.5m

Explanation:

not sure if i'm right;)

Answer:

Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

Explanation:

is this a question????...

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

Answer:

As in chemical formula? It's CH4.

Answer:

Most of us have experienced some form of electric “shock,” where electricity causes our body to experience pain or trauma. If we are fortunate, the extent of that experience is limited to tingles or jolts of pain from static electricity buildup discharging through our bodies.

When we are working around electric circuits capable of delivering high power to loads, electric shock becomes a much more serious issue, and pain is the least significant result of shock.

As electric current is conducted through a material, any opposition to the current (resistance) results in a dissipation of energy, usually in the form of heat. This is the most basic and easy-to-understand effect of electricity on living tissue: current makes it heat up. If the amount of heat generated is sufficient, the tissue may be burnt.

The effect is physiologically the same as damage caused by an open flame or other high-temperature source of heat, except that electricity has the ability to burn tissue well beneath the skin of a victim, even burning internal organs.

If atmospheric pressure is 101.3 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa, the specific gravity of the olive oil is 1.30

To calculate the specific gravity, deduct the absolute pressure at the bottom of the tank (231.3 KPa) from the atmospheric pressure (101.3 KPa). This will give you a ratio of 130KPa, then divide it by the atmospheric pressure which is the specific gravity of olive oil.

Use the formula below:

specific gravity = absolute pressure - atmospheric pressure / atmospheric pressure

Specific gravity = (231.3 KPa - 101.3 KPa) / (101.3 KPa) = 1.30

Therefore, the specific gravity of the olive oil is 1.30.

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Answer:

Newton's first law, also known as the law of inertia, states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. In other words, an object will not change its velocity on its own, it will only do so if something else pushes or pulls on it. This law is a fundamental principle of classical mechanics, and it helps to explain the behavior of objects under various conditions.

Explanation:

Answer: Newton first law states that an object at rest tends to stay at rest. an object in motion tends to stay in motion in a straight line at a constant speed until another force acts on the object.  It is also called the law of inertia

Explanation: Newton’s first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Hope this was helful

Answer: 0.0576

Explanation:

The equation for fundamental frequency for closed pipes is as follows:

f(1) = v/4L

In which v is the speed of sound (343 m/s) and L is the length of the pipe. Therefore:

528 = 343/4L

x = 0.1624

This is not the answer as you still need to subtract it from the original length.

0.220 - 0.1624 = 0.0576

Explanation:

Answer:

true . it depends tho weight depends on gravity .

Answer:

Photoelectric-type alarms aim a light source into a sensing chamber at an angle away from the sensor. Smoke enters the chamber, reflecting light onto the light sensor; triggering the alarm.

Explanation:

nfpa.org is the website with theanswer

Answer:

Explanation:true

Answer:

 α = [tex]\frac{2 \mu \ N}{m \ r}[/tex]

Explanation:

For this exercise we use Newton's equation for rotational motion

           ∑ τ = I α

the troque is

           α = Fr .r

the moment of inertia of a cylinder is

           I = ½ m r²

we substitute

         fr r = (½ m r²) α

the expression friction is

         fr = μ N

we substitute

         μ N r = ½ m r² α

         α = [tex]\frac{2 \mu \ N}{m \ r}[/tex]

Answer:

[tex]\text{(a)}\\\text{Given that current energy at starting point }h_{1} \text{ is}:\\E_{h1}=mgh+\frac{1}{2}kx^2\\\text{Plug in height and mass: }E_{h1}=10g+250(0.8)^2=10g+160=258J\\\text{Using Conservation of Energy: }E_{h1}=E_{h2} \\\therefore 258J=mg(h_{2}-h_{1})+\frac{1}{2}mv^2\\=258J=10g+v^2\\=160=v^2\\\therefore v=40m/s \text{ at }h_{2}\\[/tex][tex]\text{(b)}\\\text{Again, apply conservation of energy: } 258J=mg(h_{3}-h_{2})+\frac{1}{2}mv^2\\\text{Plugging in: }258J=16g+v^2 \\\text{ Simplify: }101.2=v^2\\\therefore v=10.05m/s \text{ at } h_{3}[/tex]

Answer:

volume=0.1m^3 and force=270N

Explanation:

Solution

Given,

mass=27kg

density=2.70×10^2=270kg/m^3

volume =?

force=?

We have,

force=m×a=27×10=270N

density =m÷v

v×d=m

v=m÷d

=27÷270

=0.1m^3

Answer:

6N

Explanation: