An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?

Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.

mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown

Answer:

v = 3.33 m/s

Explanation:

In the position of 45 degrees, all the energy of the rock is gravitational, then we have:

E = m*g*L*cos(angle)

and in the vertical position of the string, all the energy is kinetic, so we have:

E = m*v^2/2

If there is no dissipation, both energies are equal, so we have:

m*g*L*cos(45) = m*v^2/2

9.81 * 0.8 * 0.7071 * 2 = v^2

v^2 = 11.0986

v = 3.33 m/s

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B  It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

        [tex]E = \frac{\sigma}{\epsilon_o}[/tex]

Where [tex]\sigma[/tex] is the charge per unit area (Charge density )

From the question we are told that [tex]\sigma[/tex] is doubled hence the

     [tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]    

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

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Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex]     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

[tex]E_p=h\frac{c}{\lambda}[/tex]         (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

[tex]K=E_p-\Phi[/tex]     (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]

You replace in the equation (3):

[tex]K=1.95eV-1.7eV=0.25eV[/tex]

The kinetic energy of the electrons emitted by the metal is 0.25 eV

(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]

(2). The maximum kinetic energy of the electron is 0.37eV.

(1). The power of light is given as,

            [tex]P=1.4*10^{-6}W[/tex]

    Energy is given as,

             [tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]

Number of photons per second is,

                    [tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]

(2). the maximum kinetic energy of the electron is,

              [tex]K.E=E-\phi[/tex]

Where [tex]\phi[/tex] is work function.

         [tex]K.E=2.07-1.7=0.37eV[/tex]

Learn more:

https://brainly.com/question/12337396

Answer:

Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.

The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.

Explanation:

If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :

[tex]K=\dfrac{1}{2}mv^2[/tex]

m is mass of the object

We can rearrange the above equation such that,

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

Hence, this is the velocity at the end of a rollercoaster ride.

Answer:

Option B - displacement can be specified by a magnitude and a direction.

Explanation:

A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c

Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.

Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.

Looking at the options, the only one that truly justifies this definition is option B.

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.

Answer:

1.95 x 10^8 m/s.

Explanation:

Answer:

the answer is 1.95 x 10^8 m/s

Explanation:

Answer:

2025m

Explanation:

Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.

Volume of Sphere= volume of cylinder

4/3 ×π×R^3= π× r2× L

4/3 ×R^3= r^2×L

Hence

L = 3/4 × R^3/ r^2

But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}

r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}

Substituting R and r into the expression for L, we have :

L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm

202500/100= 2025m{ we divide by 100 because 100cm=1m}

Answer:

a = 17 m / s²

Explanation:

For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations

         v² = v₀² + 2a y

They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m

we clear

        a = (v² - v₀²) / 2y

we calculate

       a = (4.3² -1.2²) / 2 0.5

       a = 17 m / s²

this is the gravity of the new planet

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

Complete Question

The complete question is shown on the first uploaded image

Answer:

The experimental value of density is   [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]

Comparing it with the value of density of water ([tex]1.0*10^{3} \ kg/m^3[/tex]) we can see that the density of ice is greater

Explanation:

From the question we are told

    The mass is [tex]M = (425 \pm 25) \ g =(0.425 \pm 0.025) \ kg[/tex]

   The volume is [tex]V = (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \ m^3[/tex]

The experimental value of density is mathematically evaluated as

        [tex]\rho = \frac{M}{V}[/tex]

       [tex]\rho = \frac{0.425}{0.000405}[/tex]

       [tex]\rho = 1.05 *10^{3} \ kg/m^3[/tex]

The possible error in this experimental value of density is mathematically evaluated as

        [tex]\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}[/tex]

substituting value

         [tex]\frac{\Delta \rho}{1.05*10^{3}} = \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}[/tex]

        [tex]\Delta \rho = 101 \ kgm^{-3}[/tex]

Thus the experimental value of density is

             [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

Energy = (power) x (time)

-- For the toaster:

Power = 1.4 kW  =  1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s)  =  453,600 Joules

-- For the CFL bulb:

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s)  =  415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.  

Answer:

c. 5.50 km

Explanation:

8 min * 1h/(60 min) = 8/60 = 2/15 h

15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) =  1/240 h

8 min 15 sec = (2/15+1/240)h

40 km/h *(2/15 +1/240)h =5.50 km

Answer: 5.50 km

Explanation:

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

[tex]v_y^2=v_{oy}^2+2gh[/tex]   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]

vx is calculated by using the information about the horizontal range of the ball:

[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex]    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]

The velocity of the ball just before it touches the ground is 46.99 m/s

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

Answer:

75.0 cm

Explanation:

becouse i don,t no the right answer

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

Answer:

Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.

Answer:

a. i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Power generated is 3.6 W.

Explanation:

Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.

i.e      V = IR

i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Given that; V = 12 V and R = 40 Ohm's.

P = IV

From Ohm's law, I = [tex]\frac{V}{R}[/tex]

So that;

P = [tex]\frac{V^{2} }{R}[/tex]

   = [tex]\frac{12^{2} }{40}[/tex]

  = [tex]\frac{144}{40}[/tex]

 = 3.6 W

The power is 3.6 W.

Electrical current is defined as the flow of electric charge per unit time.