Vin Diesel jumps out of a plane. Gravity pulls on him with 184N of force and his parachute pushes him up with 82 N of force.

Net Force = (your answer)
Body Diagram or Explanation = (your answer)

Answer:

Following are the solution to this question:

Explanation:

Law:

[tex]\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}[/tex]

      [tex]= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}[/tex]

[tex]\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}[/tex]

[tex]\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\[/tex]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

[tex]\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\[/tex]

Calculating Shear Transverse:

[tex]\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}[/tex]

        [tex]= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi[/tex]

[tex]= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in[/tex]

[tex]\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\[/tex]

       [tex]= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi[/tex]

Answer:

1 m/s

Explanation:

Using law of conservation of momentum

m1v1 + m2v2 = (m1 + m2)vf , where

m1 = mass of object at rest, 20 kg

v1 = initial velocity of object at rest, 0 m/s

vf = final velocity of the bodies

m2 = mass of object in motion, 10 kg

v2 = initial velocity of object in motion, 3 m/s

On substituting, we have

(20 * 0) + (10 * 3) = (20 + 10) vf

0 + 30 = 30 vf

vf = 30 / 30

vf = 1 m/s

Therefore, the velocity of the bodies after hitting each other is 1 m/s

Answer:

The frequency [tex]f = 0.8048 \ Hz[/tex]

Explanation:

From the question we are told that

    The mass of the object is  [tex]m = 19 \ kg[/tex]

      The spring constant is  [tex]k = 486 \ N/ m[/tex]

Generally the maximum frequency is mathematically represented as  

        [tex]f = \frac{1}{2 \pi } * \sqrt{ \frac{k}{m } }[/tex]

=>     [tex]f = \frac{1}{2 * 3.142 } * \sqrt{ \frac{486}{ 19 } }[/tex]

=>     [tex]f = 0.8048 \ Hz[/tex]

Answer: 54.8

Explanation:

The centripetal force on the rock moving along the string is 104.66 N.

The given parameters;

The centripetal acceleration of the rock is calculated as follows;

[tex]\omega _f^2 = \omega _i^2 + 2\alpha (\theta)\\\\(100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} )^2 = 0 + 2\alpha (30 ^0 \times \frac{\pi \ rad}{180^0} )\\\\109.69 = 1.048\alpha \\\\\alpha = \frac{109.69}{1.048} \\\\\alpha =104.66 \ rad/s^2[/tex]

[tex]a_c = \alpha \times r\\\\a_c = 104.66 \times 1 = 104.66 \ m/s^2[/tex]

The centripetal force on the rock is calculated as follows;

[tex]F_c = ma_c[/tex]

[tex]F_c = 104.66 \times 1\\\\F_c = 104.66 \ N[/tex]

Thus, the centripetal force on the rock moving along the string is 104.66 N.

Learn more here: https://brainly.com/question/20905151

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ([tex]W[/tex]), measured in joules, is defined by the following expression:

[tex]W = F\cdot \Delta x[/tex] (1)

Where:

[tex]F[/tex] - Force, measured in newtons.

[tex]\Delta x[/tex] - Distance, measured in meters.

If we know that [tex]F = 15\,N[/tex] and [tex]\Delta x = 100\,m[/tex], then the work done by the force exerted on the object is:

[tex]W = (15\,N)\cdot (100\,m)[/tex]

[tex]W = 1500\,J[/tex]

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ([tex]a[/tex]), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

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

Where [tex]v_{o}[/tex] is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

[tex]\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t[/tex]

[tex]a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}[/tex]

If we know that [tex]\Delta x = 100\,m[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] and [tex]t = 10\,s[/tex], then the net acceleration experimented by the object is:

[tex]a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}[/tex]

[tex]a = 2\,\frac{m}{s^{2}}[/tex]

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

[tex]\Sigma F = F - f = m\cdot a[/tex] (3)

Where:

[tex]F[/tex] - External force exerted on the object, measured in newtons.

[tex]f[/tex] - Kinetic friction force, measured in newtons.

If we know that [tex]F = 15\,N[/tex], [tex]m = 6\,kg[/tex] and [tex]a = 2\,\frac{m}{s^{2}}[/tex], the kinetic friction force is:

[tex]f = F-m\cdot a[/tex]

[tex]f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)[/tex]

[tex]f = 3\,N[/tex]

The work done by friction ([tex]W'[/tex]), measured in joules, is:

[tex]W' = f\cdot \Delta x[/tex] (4)

[tex]W' = (3\,N) \cdot (100\,m)[/tex]

[tex]W' = 300\,J[/tex]

And the net work experimented by the object is:

[tex]\Delta W = 1500\,J - 300\,J[/tex]

[tex]\Delta W = 1200\,J[/tex]

By the Work-Energy Theorem we understand that change in translational kinetic energy ([tex]\Delta K[/tex]), measured in joules, is equal to the change in net work. That is:

[tex]\Delta K = \Delta W[/tex] (5)

If we know that [tex]\Delta W = 1200\,J[/tex], then the change in translational kinetic energy is:

[tex]\Delta K = 1200\,J[/tex]

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

Answer:

(1) the potential difference that stopped the proton is 1878.75 V

(2) the initial kinetic energy of the proton is 1878.75 eV

Explanation:

Given;

initial speed of the proton, v = 600,000 m/s

mass of proton, m = 1.67 x 10⁻²⁷ kg

(1) The work done in bringing the proton to rest is given as;

[tex]W = eV[/tex]

Apply work energy theorem;

[tex]K.E =W\\\\ \frac{1}{2} mv^2 = eV\\\\V = \frac{mv^2}{2e}[/tex]

where;

V is the potential difference

[tex]V = \frac{1.67\times 10^{-27} \times\ 600,000^2}{2 \ \times \ 1.6 \times 10^{-19}} \\\\V = 1878.75 \ V[/tex]

(2) the initial kinetic energy of the proton, in electron volts;

[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 1.67\times 10^{-27} \times 600,000^2 = 3.006 \times 10^{-16} \ J\\\\K.E = \frac{3.006 \times 10^{-16} \ J \ \ eV}{1.6 \times 10^{-19} \ J} = 1878.75 \ eV[/tex]

Answer:

Fr = 55 [N]

Explanation:

We must perform a free body diagram to understand the forces acting on the couch. In the attached image we see the force that Jack exerts just like that of Rob acting in the same direction.

We must perform a summation of force on the X-axis, to determine the resulting force.

∑Fx = Fr

Fr = 30 + 25

Fr = 55 [N]

Answer:

The maximum value of the magnetic field in the given wave is 6.67 nT.

Explanation:

Given;

maximum value of the electric field, E₀ = 2.0 V/m

The maximum value of the magnetic field in the given wave is calculated as;

[tex]B_o = \frac{E_o}{c}[/tex]

where;

c is the speed of light = 3 x 10⁸ m/s

[tex]B_o = \frac{E_o}{c} \\\\B_o = \frac{2}{3 \ \times \ 10^8} \\\\B_o = 6.67 \ \times \ 10^{-9} \ T\\\\B_o = 6.67 \ nT[/tex]

Therefore, the maximum value of the magnetic field in the given wave is 6.67 nT.

Answer: e: none of the above.

Explanation:

For any object in the air, the gravitational acceleration will be always -9.8m/s^2, where the negative sign is because gravity pulls the object down.

The instantaneous velocity, as a function of time, for the case of the ball, is

V(t) = (-9.8m/s^2)*t + v0

Where v0 is the velocity at which the ball is thrown up.

The velocity will be zero when the ball is at the top of its flight pat, in that point the sign of the velocity changes, it stops being positive (so the ball stops going up) and becomes negative (so the ball starts to fall down).

Now, while the instantaneous velocity can be zero during the flight, the acceleration does not, it only will be zero when the object hits the ground. Then the only correct option will be e: none of the above.

Answer:

a) Newton proposed three laws to explain the motion of things.

b) bodies are in motion the kinetic moment   is preserved,

c) entropy

Explanation:

In this exercise you are asked to explain various concepts of physics

a) Newton proposed three laws to explain the motion of things.

* The first law or inertia states that a body in an inertial system (without acceleration) remains stationary or with constant velocity if the sum of the forces is zero

* The second law states that the force (interaction) is proportional to the masses and the acceleration of the bodies

        F = m a

* The third law or action and reaction, says that the forces always go in pairs, that is, if a body interacts with a body 2, and body 2 will also interact with body 1, with the same force but with the opposite direction, each force is applied to a body

b) when bodies are in motion the kinetic moment

          L = I w

it is preserved, if it does not interact with the outside.

The quantity I called the moment of inertia is I = m r²

In this case

in the initial instant with outstretched arms

     L = I₀ w₀

in the final moment. By shrinking your arms

    [tex]L_{f}[/tex] = I_{f} w_{f}

as the body does not interact with another

     L₀ = L_{f}

     I₀ w₀ = I_{f} w_{f}

     w_{f} = [tex]\frac{I_{o} }{I_{f} }[/tex]  w₀

we substitute

      w_{f} = \frac{r_{o} }{r_{f} }^2 w₀

as the distance of the arms extended is greater than when they are retracted (r₀>  [tex]r_{f}[/tex]), the final angle speed increases

c) This is an application of the concept of entropy that establishes that disorder is the most probable state of a system, which is why the cave increases its disorder or entropy over time since by not fixing things, they are increasingly dispersed

Answer:

323.4 N

Explanation:

Given that,

The coefficient of static friction, [tex]\mu_s=0.63[/tex]

The coefficient of kinetic friction, [tex]\mu_k=0.49[/tex]

We need to find the frictional force acting on the player. As the baseball player is sliding into home base, we will use the coefficient of kinetic friction to find the frictional force.

[tex]F=\mu _kN\\\\=0.49\times 660\\\\=323.4\ N[/tex]

So, the frictional force acting on the player is 323.4 N.

Answer:

bowling was one of the first racially integrated sports because bowling alleys were primarily located in urban areas. Racial integration was inevitable since it was promoted by US Armed Forces during 1940s and its image as sport for the common man made it a choice of activity for Americans!!

Explanation:

Answer:

Just 3

Explanation:

I believe the other two are incorrect

Answer:

Even though the cross-sectional area of each capillary is extremely small compared to that of the large aorta, the total cross-sectional area of all the capillaries added together is about 1,300 times greater than the cross-sectional area of the aorta because there are so many capillaries

Explanation:

Answer:

(a) The kinetic energy at A is 6.875 J.

(b) The speed at point B is 5.26 m/s.

(c) The total work done on the particle as it moves from A to B is 0.725 J.

Explanation:

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless a change in speed occurs or the body returns to its state of rest by applying a force.

Kinetic energy is represented by the following formula:

Ec = ½ *m*v²

Where Ec is kinetic energy, which is measured in Joules (J), m is mass measured in kilograms (kg), and v is velocity measured in meters over seconds (m/s).

(a) In this case, you know:

Replacing:

Ec = ½ *0.55 kg*(5 m/s)²

and solving you get:

Ec= 6.875 J

The kinetic energy at A is 6.875 J.

(b) In this case, you know:

Replacing:

7.6 J = ½ *0.55 kg*v²

and solving you get:

[tex]v=\sqrt{\frac{7.6 J}{\frac{1}{2}*0.55 kg } }[/tex]

v= 5.26 m/s

The speed at point B is 5.26 m/s.

(c)  The total work done on the particle as it moves from A to B is the difference between the work done at the end point (point B) and the work done at the start point (point A). Considering that work is equal to kinetic energy, then:

Work= Ec at point B - Ec al point A= 7.6 J - 6.875 J

Work= 0.725 J

The total work done on the particle as it moves from A to B is 0.725 J.

Explanation:

Given that,

Distance covered, d = 40 m

Time, t = 9.5 s

Final velocity, v = 2.75 m/s

(a) Let u be the original speed of the truck. We can find it using first equation of motion.

[tex]v=u+at\\\\2.75=u+2.75\times 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s[/tex]

(b) Acceleration = rate of change of velocity

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{2.75-(-23.375)}{9.5}\\\\=2.75\ m/s^2[/tex]

So, the original speed is -23.375 and acceleration is 2.75 m/s².

Answer:

The  correct option is b

Explanation:

From the question we are told that

    The resistance of the first resistor is  [tex]R_1 = 39 \ \Omega[/tex]

     The resistance of the second resistor is  [tex]R_2 = 56 \ \Omega[/tex]

      The capacitive reactance of the first capacitor is [tex]jX_{c_1 } = 80 \ \Omega[/tex]

       The capacitive reactance of the first capacitor is [tex]jX_{c_2 } = 40 \ \Omega[/tex]

Generally given that the resistors are connected in parallel , their equivalent resistance is  

            [tex]R_e = R_1 +R_2[/tex]

=>         [tex]R_e = 39 + 56[/tex]

=>         [tex]R_e = 95 \ \Omega[/tex]

Generally given that the capacitors are connected in parallel , their equivalent capacitive reactance  is  

          [tex]jX_e = jX_{c_1} + jX_{c_2}[/tex]

=>       [tex]jX_e = 80 + 40[/tex]

=>       [tex]jX_e = 120[/tex]

Hence the impedance of the circuit is  

         [tex]Z = R_e - jX_e[/tex]        

=>      [tex]Z = 95 - j120[/tex]          

Generally from the impedance equation , the total used resistance is  

       [tex]R_e = 95 \ \Omega[/tex]

Answer:

25000N

Explanation:

distance travelled(d)=500m

force(f)=20N

Power=0.4KW=0.4×1000Watt=400watt

Then,Time(T)=?

Now,

Power(P)=W/T

400=F×D/T.(w=f×d)

or, 400=20×500/T

or, 400=10000/T

or, T=10000/400

or,T=25 sec

Therefore,time taken=25 sec

Answer:

As the height increases the pressure must increase.

Explanation:

When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression

           P = ρ g h

where rho is constant for a given fluid and h is the height measured from the surface of the fluid.

As the height increases the pressure must increase.

Answer:

The two criteria that are most essential are options C and D

Explanation:

In trying to develop a new process/procedure for the commercial/industrial production of a particular substance (in this case phosphorus pentachloride), it is important that the workers that would operate the equipment required for this new procedure are able to operate it safely, i.e without fear of injury or losing one's life. If this fear is present or any adverse effect occurs, it would not only be counterproductive but would also make workers avoid using this procedure and may be forced to revert to previous safe procedure.

It is also essential for the new procedure to be able to get the most desired product out of the reactant, meaning the rate and amount of desired product (phosphorus pentachloride) that would be obtained from a given amount of reactant must be very high (a better upgrade when compared to the previous procedure). This is going to cause for a better amount of phosphorus pentachloride produced as against the previous amounts produced from previous procedure.

From the above, it can be deduced that options C and D are two criteria that are most essential when trying to develop a novel process to produce phosphorus pentachloride.

Answer: the guy above me is correct

Explanation:

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

Answer:

number 2 down is: safetygoggles

Explanation:

no space

Lab safety personal protective equipment should always be put on when in the lab, and lab equipment are what scientists for lab work

The words for the lab safety and lab equipment puzzle are:

1. Test tube rack

2. Splash goggles

3. Fire extinguisher

4. Balance scale

5. Closed heel

6. Deluge shower

7. Waft

8. Pipette

9. Proctor

10. Meter stick

11. Bunsen burner

12. Cleanup

13. Beaker tongs

14. Measuring cylinder

15. Funnel

16. Trash bin

17. Hot plate

18. Beaker

19. Lab apron

20. Fire blanket

Learn more about lab equipment here:

https://brainly.com/question/13142279

https://brainly.com/question/13164015

https://brainly.com/question/17391361

Answer:

10 kg

Explanation:

The mass of an object does not change even if the amount of gravtiy changes.

Answer:

Because as we climb higher the amount of oxygen in the air decreases and this makes the air thinner and dryer. And at higher altitudes the blood pressure inside of our body is more than the atmospheric pressure which forces the blood to come out from openings like the nose.

Explanation:

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s[/tex]

(b) The frequency of oscillation is given by :

[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz[/tex]

Hence, this is the required solution.

Answer:

Conductors- copper, aluminum, gold, and silver.

Insulators- glass, air, plastic, rubber, and wood.

Explanation:

Answer: 5 conductors are copper, aluminum, gold, iron, and silver and 5 insulators are glass, air, plastic, rubber, and wood.