Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.

Answer:

Option 1

Explanation:

I always see animals do that

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Answer:

1.23

Explanation:

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➩ 1.23 feet

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Given :

To find :

Solution :

As it is told that it's divided into four equal pieces

Therefore,

We must divide it by 4 to get the length of each piece.

So,

[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]

Answer: True

Explanation:

[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Kinematics in real world.

So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,

P = m * V

momentum = mass * velocity

here, P= 152625 kgm/s and v= 11.1 m/s

so substituting we get as,

m = 152625 ÷ 11.1 => 13,750 kg

hence,the mass of the bus is 13,750 kg.

Answer:

[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]

Explanation:

Given

[tex]Object_1 = m_1[/tex]

[tex]Object_2 = m_2[/tex]

[tex]Distance = r[/tex]

Required

Determine the force of attraction

This is calculated as:

[tex]F = \frac{GMm}{d^2}[/tex]

Where

M = mass of object 1

m = mass of object 2

d = distance

Where G = gravitational constant

[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]

Substitute these values in

[tex]F = \frac{GMm}{d^2}[/tex]

[tex]F = \frac{6.67408 * 10^{-11} * m_1 * m_2}{r^2}[/tex]

[tex]F = \frac{6.67408 * m_1 * m_2* 10^{-11}}{r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2* 10^{-11}}{r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2}{10^{11}*r^2}[/tex]

[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]

Answer:

Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.

Explanation:

Answer:

wavelength =  0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 *  0.220 s/2

x= 162.8 m

Answer:

To find the weight of something, simply multiply its mass by the value of the local gravitational field, and you get a result in newtons (N). For example, if your mass is 50 kg (about 110 pounds), then your weight is (50) (9.8). The point that must be overwhelmingly emphasized is that weight is a force.

Explanation:

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

A) The change in the temperature of the gas is; ΔT = 139.5 K

B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J

C) The change in pressure of the gas is; ΔP = 2899.5 Pa

We are given;

Volume of Monatomic ideal gas; V = 1.2 m³

Amount of heat added; Q = 5.22 × 10³ J

number of moles; n = 3

A) To calculate the change in temperature of the monatomic idea gas, we will use formula;

Q = ³/₂nRΔT

Where R is a constant = 8.314 J/mol.K

ΔT is the change in temperature

Making ΔT the subject of the formula;

ΔT = ²/₃(Q/(nR))

ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)

ΔT = 139.5 K

B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;  

ΔU = Q

Thus;

change in internal energy; ΔU = 5.22 × 10³ J

C) The change in pressure will be calculated from the formula;

ΔP = (n*R*ΔT)/V

ΔP = (3 * 8.314 * 139.5)/1.2

ΔP = 2899.5 Pa

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Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

Answer:

The final velocity of the car is 16.893 m/s

The final velocity of the truck is 4.393 m/s

Explanation:

Given;

mass of the car, m₁ = 715 kg

mass of the truck, m₂ = 1490 kg

initial velocity of the car, u₁ = 0

initial velocity of the truck, u₂ = 12.5 m/s

let the final velocity of the car, = v₁

let the final velocity of the truck, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂

18625 = 715v₁ + 1490v₂ -----equation (1)

Apply one-directional velocity formula;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 12.5 + v₂

v₁ = 12.5 + v₂

Substitute v₁ into equation (1)

18625 = 715(12.5 + v₂) + 1490v₂

18625 =8937.5 + 715v₂ + 1490v₂

18625 - 8937.5 = 715v₂ + 1490v₂

9687.5 = 2205v₂

v₂ = 9687.5 / 2205

v₂ = 4.393 m/s

solve for v₁

v₁ = 12.5 + v₂

v₁ =  12.5 + 4.393

v₁ = 16.893 m/s

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.

Answer:

sorry I wish I could it help you

Answer:

the answer is a time your welcome

Answer:

(1)

Explanation:

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

The distance should be 480.2 m

Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s

[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]

2x = 960.4

x = 480.2 m

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Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.

• w = weight = 319 N

• n = normal force

• p = pushing force = 485 N

• f = friction = µ n, where µ is the coefficient of static friction

The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)

• vertical:

p sin(-35°) + n - w = 0

and solving for n,

- (485 N) sin(35°) + n - 319 N = 0

n ≈ 597 N

• horizontal:

p cos(-35°) - f = m a

where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get

p cos(35°) - µ n = (w / g) a

(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a

a ≈ (12.2 - 18.3 µ) m/s²

(a) If µ = 0.57, then the net acceleration on the box is

a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²

so that the time t required to move the box 4 m is

4 m = 1/2 a t ²

t ≈ √((8 m) / (1.75 m/s²))

t ≈ 2.14 s

(b) The box does not move.

If µ = 0.75, then

a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²

but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.

(a) The time taken to move the box 4 meters is 2.14 s.

(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

The given parameters;

The mass of the books is calculated as;

[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]

The normal force on the box is calculated as follows;

[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]

The frictional force when the coefficient of friction is 0.57;

[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]

The time taken to move the box 4 meters is calculated as;

[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]

(b) The frictional force when the coefficient of friction is 0.75;

[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]

Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

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Answer:

false

Explanation:

Answer:

$4.00

Explanation:

The cost per pound is $4.00, $11 divided by 2.75 is 4.

Answer:

Th cost per pound is $4.00.

Explanation:

11/2.75= 4

Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.

Explanation:

Type of meter would be connected on both sides of a resistor in a circuit is a a voltmeter , which measures potential difference .

A voltmeter is an instrument used for measuring the potential difference , or voltage between two points in an electrical circuit .

A voltmeter is always attached in a series combination and an ammeter (which measures current in a circuit ) always attached in parallel combination with the circuit.

Since , in question it is given that a meter would be connected on both sides of a resistor in a circuit that means it must be a series combination

hence , correct answer is B) a voltmeter , which measures potential difference .

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#SPJ3

Answer:

option 3

Explanation:

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Answer:

v = 12.8 m/s

Explanation:

       [tex]p_{ox} = p_{fx} (1)[/tex]

       [tex]p_{oy} = p_{fy} (2)[/tex]

       [tex]p_{oy} = m_{t} * v_{ot} = 3520 kg* v_{ot} (3)[/tex]

       [tex]p_{fy} = m_{f} * v_{fy} = 5000 kg* 9.8 m/s * sin 66.9 = 45080 kg*m/s (4)[/tex]

       [tex]v_{ot} =\frac{45080kg*m/s}{3520kg} = 12.8 m/s (5)[/tex]

Answer:

The magnitude of the bugs displacement is 3.87 m

Explanation:

An illustrative diagram for the scenario is given in the attachment below.

In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is

x² = 2.25² + 3.15²

x² = 5.0625 + 9.9225

x² = 14.985

x = √14.985

x = 3.87 m

Hence, the magnitude of the bugs displacement is 3.87 m.