A football team lost 3 yards on their first play. On the second play, 10 yards were gained. On the third play the quarterback got tackled behind the line of scrimmage and lost 6 yards. What was their net gain or loss after the three plays?​

This question is incomplete, the complete question is;

The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .

At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.

The phase constant φ is given by tanφ =:

A) ωv₀ /y₀    

B) ωv₀ y₀  

C) v₀ /ωy₀  

D) y₀ /ωv₀      

E) ωy₀ /v₀

Answer:

E) ωy₀ /v₀

Explanation:

Given that;

displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)

we differentiate the given equation with respect to time

d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )

v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )

v₀ = -ym ωcos (-φ)  ......... lets leave thisas equ 1

At t = 0, x = 0

the displacement of the wave is

y(0,0) = ym sin (k(0) - ω(0) - φ)

y₀ = ym sin(-φ) ..............let this be equ 2

y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))

(tanφ)/ω = y₀/v₀

tanφ = y₀ω/v₀

therefore the required value is y₀ω/v₀

option (E).  

Answer:

The acceleration of the car is 10.16m/s²

Explanation:

Given parameters:

  Initial velocity = 8.77m/s

   Final velocity = 47.8m/s

   Time duration  = 3.84s

Unknown:

Acceleration of the car = ?

Solution:

To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;

     Acceleration  = [tex]\frac{V - U}{T}[/tex]

V is the final velocity

U is the initial velocity

T is the time taken

  Input the parameters and solve for acceleration;

      Acceleration  = [tex]\frac{47.8 - 8.77}{3.84}[/tex]   = 10.16m/s²

The acceleration of the car is 10.16m/s²

Answer:

θ = 33.0705°

Explanation:

The angle between the two vectors is given by the formula;

Cos θ = (F1 • F2)/(|F1| × |F2|)

We are given;

F1 = 8.92i + 17.37j

F2 = 12.44i + 7.11j

Thus;

Cos θ = [(8.92i + 17.37j) • (12.44i + 7.11j)]/[√(8.92² + 17.37²) × √(12.44² + 7.11²)]

Cos θ = (110.9648 + 123.5007)/(19.5265 × 14.3285)

Cos θ = 0.8380

θ = cos^(-1) 0.8380

θ = 33.0705°

Answer:

11760 joules

Explanation:

Given

Mass (m) = 75kg

Height (h) = 16m

Required

Determine the increment in potential energy (PE)

This is calculated as thus:

PE = mgh

Where g = 9.8m/s²

Substitute values for m, g and h.

P.E = 75 * 9.8 * 16

P.E = 11760 joules

The potential energy of the man in the fifth floor of the building has increased by 11760J.

Given the data in the question;

Potential energy; [tex]P_E =\ ?[/tex]

Potential energy is the energy possessed by a particle due to its position relative to other particles. It is expressed as:

[tex]P_E = mgh[/tex]

Where m is mass of the particle, h is its height above ground level and g is acceleration due to gravity( [tex]g = 9.8m/s^2[/tex] ).

We substitute our values into the equation

[tex]P_E = 75kg\ *\ 9.8m/s^2\ *\ 16m\\\\P_E = 11760kg.m/s^2\\\\ P_E = 11760J[/tex]

Therefore, the potential energy of the man in the fifth floor of the building has increased by 11760J.

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The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.

Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.

Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.

The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.

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Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

where

f is the force

m is the mass

From the question

f = 20 N

m = 10 kg

We have

[tex]a = \frac{20}{10} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:.

Explanation:.

Answer:

1>500gf

1>300gf

its answer

Explanation:

because of the acceleration due to gravity is constant which is 9.8

Answer:

TRUE

Explanation:

CUZ , IT DOSEN'T AFFECT

Mass does not affect the speed of falling objects, assuming there is only gravity acting on it. Both bullets will strike the ground at the same time.

Answer:

there is only a small amount of gravity present.

Explanation:

this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.

Answer:

Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

Answer:

[tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Explanation:

Given that,

The radius of a sphere is (6.45 ± 0.30) cm

Mass of the sphere is (1.79 ± 0.08) kg

Density = mass/volume

For sphere,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]

We can find the uncertainty in volume as follows :

[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]

Uncertainty in mass,

[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]

Now, the uncertainty in density of sphere is given by :

[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]

Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Answer:the answer is 3

Explanation:

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN

Answer:

The value  is  [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

Explanation:

From the question we are told that  

   The speed of the airplane from the north is [tex]\frac{dN}{dt}  =  -181 \  km /hr[/tex]

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  [tex]\frac{dE}{dt}  =  -278 \  km/hr[/tex]

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  [tex]N  =  30 \  km[/tex]

   The distance of the westbound plane is  [tex]E =  15 \  km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    [tex]R^2  = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     [tex]2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}[/tex]

Here

     [tex]R = \sqrt{N^2 + E^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R =33.54 \ m [/tex]

    [tex]2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)[/tex]

=>   [tex] 67.08 * \frac{dR}{dt} =  -19200[/tex]

=>   [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

The rate of change of the distance between the planes is 286.23 km/hr.

The given parameters;

The distance between the two planes is calculated by applying Pythagoras theorem as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The rate of change of the distance between the planes is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the rate of change of the distance between the planes is 286.23 km/hr.

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Answer:

22 crates

Explanation:

Power = Force ×Distance/time taken

Power = m×a×d/t

Power = 30×9.81×0.9/60 (1min was converted to second)

Power = 264.87/60

Power = 4.4145Watts

If my average power output us 100aw, then;

Number of crate to load = average power/4.4145

Number of crates to load = 100/4.4146

Number of crates to load = 22.6

Hence I will have to load about 22 crates onto the truck in 1 minute for my average power output in lifting the crates to be 100 W.

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below  

Answer:

40N

Explanation:

trust

Answer:

The moment of inertia of the object is 17.276 kilogram-square meters.

Explanation:

According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ([tex]K_{R}[/tex]), measured in joules, is represented by the following formula:

[tex]K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}[/tex] (Eq. 1)

Where:

[tex]I_{G}[/tex] - Moment of inertia with respect to center of mass, measured in kilogram-square meters.

[tex]\omega[/tex] - Angular speed, measured in radians per second.

Now we clear the moment of inertia:

[tex]I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}[/tex]

If we know that [tex]K_{R} = 16\,J[/tex] and [tex]\omega \approx 1.361\,\frac{rad}{s}[/tex], then the moment of inertia of the object is:

[tex]I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]I_{G} =17.276\,kg\cdot m^{2}[/tex]

The moment of inertia of the object is 17.276 kilogram-square meters.

The moment of inertia of the object will be "17.276 kg/m²".

Rotational Kinetic energy, [tex]K_R[/tex] = 16 J

Angular speed, ω = 1.361 rad/s

By using the Rotation Physics, the relation will be:

→ [tex]K_R[/tex] = [tex]\frac{1}{2}[/tex] × [tex]I_G[/tex] × ω²

the,

The moment of inertia be:

→ [tex]I_G[/tex] = [tex]\frac{2\times K_R}{\omega^2}[/tex]

By substituting the values, we get

       = [tex]\frac{2\times 16}{(1.361)^2}[/tex]

       = [tex]\frac{32}{(1.361)^2}[/tex]

       = 17.276 kg.m²

Thus the above answer is correct.  

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Answer:

ggggggggggggggggggggggggggggg

Explanation:

Answer:

The volume of the sample of gold is

16.51 [tex]cm^{3}[/tex]

Explanation:

The formula for density is:

D= [tex]\frac{M}{V}[/tex].

where:

D is density,

M is mass, and

V is volume.

Rearrange the density formula to isolate volume.

V= [tex]\frac{M}{D}[/tex]

V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]

V= 318.97∅ ×  [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.

V= 16.51 cm³ Au.

Answer:

10 seconds

Explanation:

As it starts from rest, then u=0

and by III rd equation of motion:

d. 49.0 m/s

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Answer:

(a) 102 cm/s

(b) 0.490 cm²

Explanation:

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

Answer:

I think it would be 50 I am not really sure

Explanation:

I think you would have to divid 1000 by 20 Again I'm not sure

Answer:

5400 N

Explanation:

f=ma

f= 1200*4.5

f=5400N