NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Answer:

Acceleration and mass.

Explanation:

Depends on how heavy the car is and how fast it is going.

F=MA

We are given:

F = 300N

acceleration (a) = 0.750 m/s²

Solving for the mass of the object:

We know from the second equation of motion:

F = ma

replacing the variables

300 = m * 0.75

m = 300 * 100/ 75

m = 300 * 4/3

m = 100 * 4

m = 400 kg

Answer: Temperature

Explanation:

The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer.

"The kinetic energy of an object is the energy that it possesses due to its motion."

"The product of the half of the mass of each gas molecule and the square of RMS speed."

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Answer: sorry here’s the answers, I didn’t feel like typing it all

Explanation:

The correct answer to the 5 questions are;

1) Option A; The current stays the same throughout the circuit.

2) Option A;  The current will stay the same at 1 amp after passing through both resistors.

3) Option C; R1: 12 V, R2: 12 V

4) Option A; I1 is equal to I2

5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.

1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.

2) Formula for current is;

I = V/R

We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.

Thus; Current = 10/10 = 1 A.

The current will stay same at 1 A after passing through both resistors.

3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.

Option C is correct

4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.

5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.

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Given parameters:

Mass of puck = 162g  = 0.162kg (1000g = 1kg)

Force exerted on puck = 171N

Final velocity  = 42.3m/s

Unknown

A.  time of the acceleration

B. radius of the curve?

Solution:

A. time of the acceleration

the initial velocity of the puck = 0m/s

    We know that;

              Force  = mass x acceleration

      Acceleration  = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]

       Acceleration  = [tex]\frac{42.3 - 0}{t}[/tex]

  So force  = mass x [tex]\frac{42.3 }{t}[/tex]

 Input the parameters and solve for time;

             171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]

             171 = [tex]\frac{6.85}{t}[/tex]  

                 t = [tex]\frac{6.85}{171}[/tex]   = 0.04s

The time of acceleration is 0.04s

B. radius of the curve;

      to solve this, we apply the centripetal force formula;

               F  = [tex]\frac{mv^{2} }{r}[/tex]

  where;

      F is the centripetal force

     m is the mass

      v is the velocity

      r is the radius

               Since the force exerted on the puck is 151;

      input the parameters and solve for r²;

             151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]

               151r  = 0.162 x 42.3²

                r  = 1.92m

The radius of the circular curve is 1.92m

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:

[tex]E = U_{g} + K[/tex] (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.

[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.

If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:

[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 498556.296\,J[/tex]

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)

Where:

[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.

[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]

[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.

[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.

If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:

[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]

[tex]W_{loss} = 125960.4\,J[/tex]

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)

[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]

Where:

[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.

[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.

[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.

[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)

[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]

[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]

[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]

If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:

[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]

[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

630 kilometers for the whole journey

Answer

No dan should reduce his discretionary apendings.

Explanation:

Answer:

The car goes, x = 200 [m]

Explanation:

In order to solve this problem we must use the following kinematic equation.

[tex]v_{f} ^{2} =v_{i} ^{2} - (2*a*x)[/tex]

where:

Vf = final velocity = 0

Vi = initial velocity = 40 [m/s]

a = desacceleration = 4 [m/s^2]

x = distance traveled [m]

Note: The negative sign of the above equation indicates that the vehicle is slowing down

Now replacing:

0 = (40)^2 - (2*4*x)

(0 - 40^2) = - 8*x

x = 200 [m]

Answer:

Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0), then that object will have zero acceleration. That doesn't necessarily mean the object is at rest, but it means that the velocity is constant.

Explanation:

According to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.

Newton's first law of motion states that an object in state of rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

This first is also known as the law of inertia because it is the reluctance of an object in motion to stop moving or reluctance of an object at rest to start moving which depends on the mass of the object.

Thus, we can conclude that according to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.

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Answer:

F = 1280 N

Explanation:

Given that,

Acceleration experienced by a space shuttle, a = 32 m/s²

Mass of the astronauts, m = 40 kg

We need to find the force experienced by the astronaut.

We know that the net force is equal to the product of acceleration and its mass. So,

F = ma

F = 40 kg × 32 m/s²

So,

F = 1280 N

So, 1280 N of force is experienced by the Astronaut.

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Answer:

k = 930 N / m

Explanation:

For this problem we will solve it in parts, let's start with the conservation of mechanical energy

Starting point. Lower

          Em₀ = [tex]K_{e}[/tex] + U₁

Final point. Higher

         [tex]Em_{f}[/tex]= U₂

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          ½ m v² + mg y₁ = m g y₂

where y₁ is the initial height of y1 = -0.5 m and y² the final height y² = 15 m

Let's find speed when getting out of the sling

         v = √ (2g (y₂-y₁))

let's calculate

         v = √[2 9.8 (15 - (-0.5))]

         v = 17.43 m / s

Now we can use Newton's second law.

The force applied by the sling is in the direction of movement (inclined) and the weight is in the vertical direction.

X axis

         Fₓ = m aₓ

in the problem they indicate that the direction of the velocity at the end of the sling is the same direction of the force,  

         F_{e} cos θ = m a cos θ

let's replace the elastic force

        k Δx = m a

Y axis

        F_{y} - W = m a_{y}

         k Δx sin θ - m g = m a sin θ

let's write

              k Δx = m a        (1)

            k Δx sin θ - m g = m a sin θ

Now let's use kinematics to find the acceleration in the sling, the direction of these accelerations ta in the direction of elongation

           v² = v₀² +2 a Δx

as the system starts from rest v₀ = 0

           a = v² / 2Δx

           a = 17.43² / (2  1.4)

           a = 108.5 m / s²

we substitute in equation     1

            k = m a / Δx

            k = 12 108.5 / 1.4

            k = 930 N / m

Answer:

v = 190 m/s

Explanation:

Given that,

Initial velocity of a car, u = 10 m/s

Distance, d = 500 m

Time, t = 5 s

We need to find the final velocity of the car. Firstly we can find the acceleration of the car using second equation of motion as follows :

[tex]d=ut+\dfrac{1}{2}at^2\\\\500=10\times 5+\dfrac{a}{2}\times 5^2\\\\500=50+\dfrac{25a}{2}\\\\450=\dfrac{25a}{2}\\\\a=\dfrac{450\times 2}{25}\\\\a=36\ m/s^2[/tex]

let v is the final velocity. using First equation of motion to find a as follows :

v=u+at

v=10+36(5)

v=190 m/s

So, the final velocity of the car is 190 m/s.

Answer:

Can't really answer that for now. More context please?

Explanation:

I will answer it in a comment when you give some context.

Answer:

D.Speed

Explanation:

The speed of an object is the distance the object travels in one unit of time.

Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.

The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

Thus, Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.

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Answer:

oh I know that sounds good to me

Answer:

499 seconds or a little over 8 minutes

Explanation:

~150 million km/ 300,000 km per second = 500 seconds

Answer:

It would take 5 seconds

Explanation:

I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.

V = d / t

34 = 170 / t

34 x t = 170

34t = 170

t = 170 / 34

t = 5

Answer:

5m

Explanation:

Answer:

5

Explanation:

because that's the average of how far it when the most

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

  A

2

A

Explanation:

From the question the data given for Error (mV) is -15

-15.17

8.67

-13.74

-20.69

-6.96

-1.36

-2.96

-9.26

3.11

-14.12

6.39

-14.77

Generally

The null hypothesis is [tex]H_o : \mu = 0[/tex]

The alternative hypothesis is [tex]H_a : \mu \ne 0[/tex]

The sample size is n = 13

Here [tex]\mu[/tex] represents the true error bias (i.e population error bias)

Generally the sample error bias is mathematically represented as

[tex]\= x = \frac{ \sum x_i}{n}[/tex]

=> [tex]\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}[/tex]

[tex]\= x = -7.37[/tex]

Generally the standard deviation is mathematically represented as

[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }[/tex]

=> [tex]\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }[/tex]

=> [tex]\sigma = \sqrt{ 119.385}[/tex]

=> [tex]\sigma = 10.926[/tex]

Generally the test statistics is mathematically represented as

[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

=> [tex]t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }[/tex]

=> [tex]t = -2.838[/tex]

Generally the p-value is mathematically represented as

[tex]p-value = 2 P(t < -2.432)[/tex]

From the z-table  [tex]P(t < -2.432) =  0.0075 [/tex]

So  [tex]p-value  =  2* 0.0075 [/tex]

=>  [tex]p-value  = 0.015 [/tex]

So given that  p-value is  less than the [tex] \alpha = 0.05[/tex] then we reject the null hypothesis and conclude that the oscilloscope has an error bias

Answer:

The amount of heat required is [tex]H_t = 1.37 *10^{6} \ J [/tex]

Explanation:

From the question we are told that

The mass of water is [tex]m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g[/tex]

The temperature of the water before drinking is [tex]T_w = 3.8 ^oC[/tex]

The temperature of the body is [tex]T_b = 36.6^oC[/tex]

Generally the amount of heat required to move the water from its former temperature to the body temperature is

[tex]H= m_w * c_w * \Delta T[/tex]

Here [tex]c_w [/tex] is the specific heat of water with value [tex]c_w = 4.18 J/g^oC [/tex]

So

[tex]H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)[/tex]

=> [tex]H= 7.8 *10^{4} \ J [/tex]

Generally the no of mole of sweat present mass of water is

[tex]n = \frac{m_w}{Z_s}[/tex]

Here [tex]Z_w[/tex] is the molar mass of sweat with value

[tex]Z_w = 18.015 g/mol[/tex]

=> [tex]n = \frac{5.7 *10^2}{18.015}[/tex]

=> [tex]n = 31.6 \ moles [/tex]

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

[tex]H_v = n * L_v[/tex]

Here [tex]L_v[/tex] is the latent heat of vaporization with value [tex]L_v = 7 *10^{3} J/mol[/tex]

=> [tex]H_v = 31.6 * 7 *10^{3} [/tex]

=> [tex]H_v = 1.29 *10^{6} \ J [/tex]

Generally the overall amount of heat energy required is

[tex]H_t = H + H_v[/tex]

=> [tex]H_t = 7.8 *10^{4} + 1.29 *10^{6}[/tex]

=> [tex]H_t = 1.37 *10^{6} \ J [/tex]

Answer:

2.5

Explanation:

Answer:

Explanation:

Newtons third law says an applied force will produce an equal but opposite force.

[tex]F_A_B =-F_B_A[/tex]