1.A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water

2.What is the horizontal distance the boy in # 1 travels while in the air ? ​

Answer:

f

Explanation:

f

Answer:

Due to the effect of elevation on the experimental group the participants decided to try to help the research assistant who was having opening one of her files to finish the study. Schnall, Roper, and Fessler were able to conclude that happiness associated with a feeling of elevation can lead to more altruism or helping behaviors.

Explanation:

100 percent on edge

The effect of elevation on the experimental group was they show more urge of being altruistic and feeling happiness and satisfaction associated with elevation.

Altruism:

It is a practice in which a person help others without any selfishness, the person just want to help.

Witnessing someone's altruistic behavior, make other to feel good and a urge of being Altruistic, this is known as elevation.

After watching elevating Oprah video, the test group help the research assistant who was having trouble in opening file.

Therefore, the effect of elevation on the experimental group was they show more urge of being altruistic and feeling happiness and satisfaction associated with elevation.

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if you stroke it an iron nail with a bar magnet the nail will become a permanent or long lasting magnet.

Hope it's perfect for you.

This question is incomplete, the missing diagram is uploaded along this Answer below.

Answer:

a) the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight is - 3.935 lb-ft

Explanation:

Given the data in the question;

(a) determine the work done on the cart by the spring

we calculate the work done on the cart by the spring as follows;

[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )

where k is spring constant ( 3 lb/in )

we substitute  

[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )      

[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )

[tex]W_{spring}[/tex] = 1/2 × 3( 39 )

[tex]W_{spring}[/tex] = 58.5 lb-in

we convert to pound force-foot

[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft

[tex]W_{spring}[/tex] = 4.875 lb-ft

Therefore, the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight

work done by its weight;

[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )        

we substitute in of values from the image below;

[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )  

[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13

[tex]W_{gravity}[/tex] = -47.1  lb-in

we convert to pound force-foot

[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft

[tex]W_{gravity}[/tex] = - 3.935 lb-ft

Therefore, the work done on the cart by its weight is - 3.935 lb-ft

a) the work done on the cart by the spring is 4.875 lb-ft.

b) the work done on the cart by its weight is - 3.935 lb-ft.

a. The work done on the cart by the spring is

= 1/2 × 3( (-8)² - (5)² )      

= 1/2 × 3( 64 - 25 )

= 1/2 × 3( 39 )

= 58.5 lb-in

Now we have to convert to pound force-foot

So,

= 58.5 × 0.0833333 lb-ft

= 4.875 lb-ft

b) Now

work done by its weight;

= -mgsin∅( x₂ - x₁ )        

So,

= -14 × sin(15°)( 5 - (-8) )  

= -14 × 0.2588 × 13

= -47.1  lb-in

Now we convert to pound force-foot

= -47.1 × 0.0833333 lb-ft

= - 3.935 lb-ft

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Answer:

1- water

2- fat

3- carbohydrates

4- vitamins

5- minerals

Answer:

B

Explanation:

Motion is movement, the teacher's movement is motion

Answer:

A Polarizing sheet transmits only the component of light polarized along a particular direction and absorbs the component perpendicular to that direction.

Consider a light beam in the z direction incident on a Polaroid which has its transmission axis in the y direction. On the average, half of the incident light has its polarization axis in the y direction and half in the x direction. Thus half the intensity is transmitted,and the transmitted light is linearly polarized in the y direction.

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

Answer:

v= 20.8 m/s

Explanation:

       [tex]v_{f} = v_{o} + a*t = v_{o} + g*t (1)[/tex]

        [tex]\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)[/tex]

       [tex]v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)[/tex]

       [tex]v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)[/tex]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

Answer:

Density of liquid

Depth of liquid

Acceleration due to gravity

Answer:

Please see below as the answer is self explanatory.

Explanation:

The distance travelled by toy car B using the same fully charged battery as car A is 875 m

E = fd

E = 35 × 2000

E = 70000 J

E = fd

70000 = 80 × Distance of car B

Divide both sides by 80

Distance of car B = 70000 / 80

Distance of car B = 875 m

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#SPJ1

Answer:

Digestive system

Explanation:

ulcer affect anywhere in the digestive system

Digestive system.

Since the acids in your food break down with the chemicals in your stomach, it can give you heartburn and also, ulcer disease happens in your stomach so the only correct answer would be Digestive System. I would like to say that the person with the profile name BigPapa who commented on my answer deserves a lot of credit, and thanks if you see this.

-R3TR0 Z3R0

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

[tex]F_x=kx[/tex]

Substitute the values

[tex]6500=1300x[/tex]

[tex]x=\frac{6500}{1300}=5[/tex]m

Work done due to friction force

[tex]W_f=fscos\theta[/tex]

We have [tex]\theta=180^{\circ}[/tex]

Substitute the values

[tex]W_f=50\times 5cos180^{\circ}[/tex]

[tex]W_f=-250J[/tex]

Initial kinetic energy, Ki=0

Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\

Initial elastic potential energy

[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]

[tex]U_{el,1}=16250J[/tex]

Final elastic energy,[tex]U_{el,2}=0[/tex]

Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]

Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]

Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]

Using work-energy theorem

[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]

Substitute the values

[tex]0+0+16250-250=30v^2+1470+0[/tex]

[tex]16000-1470=30v^2[/tex]

[tex]14530=30v^2[/tex]

[tex]v^2=\frac{14530}{30}[/tex]

[tex]v=\sqrt{\frac{14530}{30}}[/tex]

[tex]v=22m/s[/tex]

Answer:

Thomas Edisons most famous invention was the phonograph

Thomas Edison announces his invention of the phonograph, a way to record and play back sound. Edison stumbled on one of his great inventions—the phonograph—while working on a way to record telephone communication at his laboratory in Menlo Park, New Jersey.

Explanation:

Hope I helped

Answer:

megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter

Explanation:

plz mark brainlest

Answer:

2.5 cm

Explanation:

Using the relation :

Refractive index = Real Depth / Apparent depth

Refractive index = 1.6

Real depth = 4cm

Virtual depth = apparent depth = x

1.6 = 4cm / x

1.6x = 4

x = 4 / 1.6

x = 2.5

Hence, virtual depth = 2.5cm

Answer:

Explanation:

Suppose that the turbines of a coal-fired plant are driven by hot gases at a temperature of 886 K. the temperature of the exhaust area is only 305 K,  the efficiency of this heat engine

Answer:

the force acting on the car is 3600 N

Explanation:

The computation of the force acting on the car is shown below:

As we know that

Force = mass × acceleration

= 1200 kg × 3.0 ms/^2

= 3600 N

hence, the force acting on the car is 3600 N

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

Answer:

Halved

Explanation:

F=ma

Let case 1 (original) be:

[tex]F_{1}=m_{1} a_{1} \\[/tex]

Case 2 (new) be:

[tex]F_{2}=m_{2} a_{2}[/tex]

Mass is double:

[tex]m_{2}= 2m_{1}[/tex]

Force kept the same:

[tex]F_{1} =F_{2}[/tex]

Combine the equation and gives:

[tex]\frac{F_{1} }{F_{2}} =\frac{m_{1} a_{1} }{m_{2}a_{2} }\\\frac{F_{1} }{F_{1}} =\frac{m_{1} a_{1} }{2m_{1}a_{2} }\\1=\frac{a_{1} }{2a_{2} }\\a_{2}=\frac{1}{2} a_{1}[/tex]

Acceleration is halved

Answer:

Explanation:

A)

Density= charge/total volume .......eqn(1)

But volume= 4/3πr^3

r= radius= 20 cm= 0.20m

If we substitute into the volume equation, we have

volume= 4/3 * 3.142 *( 0.20)^3

= 0.0335 m^3

The volume= 0.0335 m^3

Charge=71 nC= 71×10^-9

If we substitute into eqn(1) we have

Density= (71 *10^-9C )/0.0335

= 2.11µc/m^3

B) charge enclose= Density × volume

spheres of radii are

5cm

10 cm

20 cm

Volume for 5cm

V= 4/3 * 3.142 *( 0.05)^3 = 0.0005237 m^3

charge enclose=2.11µc/m^3×0.0005237

charge enclose= 2.110 nC

Volume for 10cm

V= 4/3 * 3.142 *( 0.10)^3 = 0.004189 m^3

charge enclose= 2.11µc/m^3 ×0.004189

=8.9 nC

Volume for 20cm

V= 4/3 * 3.142 *( 0.20)^3 = 0.0335 m^3

charge enclose= 71nC

Answer:

Explanation:

The magnitude of the force is known as the resultant.

R = √Fx²+Fy²

Fx = 80cos 20 + 40cos70

Fx = 80(0.9397)+40(0.3420)

Fx = 75.176 + 13.68

Fx = 88.856N

Fy = 80sin 20 + 40sin70

Fy = 80(0.3420)+40(0.9397)

Fy = 27.36 + 37.588

Fy = 64.948N

R = √88.586²+64.948²

R = √7,847.48+4,218.24

R = √12,065.72

R = 109.5

R = 110N

Hence the magnitude of the forces is 110N

Answer:

the force needed to give the truck the acceleration is 29,760 N.

Explanation:

Given;

mass of truck, m = 4800 kg

acceleration of the truck, a = 6.2 m/s²

The force needed to give the truck the acceleration is calculated as;

F = ma

F = 4800 x 6.2

F = 29,760 N

Therefore, the force needed to give the truck the acceleration is 29,760 N.

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

(a)  The distance from the uranium atom at which an electron will be in equilibrium is [tex]2.04 \times 10^{-8} \ m[/tex]

(b) The magnitude of the force on the electron from the uranium ion is [tex]3.46 \times 10^ 6 \ N[/tex]

The given parameters:

The force on the electron due to uranium ion at distance r is calculated as follows;

[tex]F _1 = \frac{Kq_1^2}{r^2} \\\\F_1 = \frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{r^2} \\\\F_1 = \frac{2.3 \times 10^{-28}}{r^2}[/tex]

The force on the electron due to uranium ion at distance less than 61.10 nm.

R = 61.10 nm - r

[tex]F_2 = \frac{9\times 10^9 \times (3.2 \times 10^{-19})^2}{(61.1 \times 10^{-9} \ - \ r)^2} \\\\\F_2 = \frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}[/tex]

At equilibrium, the force between the electron and ions will be equal.

[tex]\frac{9.216 \times 10^{-28}}{(61.1 \times 10^{-9} \ - \ r)^2}= \frac{2.3 \times 10^{-28}}{r^2}\\\\\frac{4}{(61.1 \times 10^{-9} \ - \ r)^2} = \frac{1}{r^2} \\\\4r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2^2r^2 = (61.1 \times 10^{-9} \ - \ r)^2\\\\2r = 61.1 \times 10^{-9} \ - \ r\\\\3r = 61.1 \times 10^{-9} \\\\r = \frac{61.1 \times 10^{-9}}{3} \\\\r = 2.04 \times 10^{-8} \ m[/tex]

The magnitude of the force on the electron from the uranium ion is calculated as follows;

[tex]F = \frac{kq_1^2}{r^2} \\\\F = \frac{9\times 10^9 \times 1.6\times 10^{-19}}{(2.04 \times 10^{-8})^2} \\\\F= 3.46 \times 10^6 \ N[/tex]

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