On a dry day, the temperature in Boulder (altitude: 5330’) is 40F. What is the temperature (in F) on nearby Bear Peak (altitude: 8460’)?

Answer:

R= 78.32 N

Explanation:

Given that

Acceleration ,a= 2.05 m/s²

Mass , m = 6.5 kg

The force due to acceleration

F= mass x Acceleration

F=  ma

F= 6.5 x 2.05 N

F= 13.32 N

The force due to weight

F' = m g

F' = 6.5 x 10 N             ( take g= 10 m/s²)

F'= 65 N

Therefore the net total force will be summation of force due to weight and force due to acceleration

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

The force that the wire exerts on the instrument is R= 78.32 N

Calculation of force:

Given that

Since The force due to weight

[tex]F= ma\\\\F= 6.5 \times 2.05 N\\\\F= 13.32 N[/tex]

Now

[tex]F' = m g\\\\F' = 6.5 \times 10 N ( take\ g= 10 m/s^2)[/tex]

F'= 65 N

Now

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

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Answer:

11.245 m

Explanation:

The vertical component of the initial velocity v0 is

[tex]v_v = v_0sin58^0 = 0.848v_0[/tex]

This makes the ball reach a maximum height of 8m. If we apply the conservation law of mechanical energy, its kinetic energy is converted to potential energy when it travels to the maximum height

[tex]E_p = E_k[/tex]

[tex]mgh = mv_v^2/2[/tex]

where m is the mass and h = 8 m is the maximum vertical distance traveled, g = 9.81m/s2 is the gravitational acceleration

we can divide both sides by m

[tex]gh = v_v^2/2[/tex]

[tex](0.848v_0)^2 = 2gh = 2*9.81*8 = 156.96[/tex]

[tex]0.848v_0 = \sqrt{156.96} = 12.53[/tex]

[tex]v_0 = 12.53 / 0.848 = 14.77 m/s[/tex]

So if the ball is directed fully upward at v0 speed then we can apply the same equation to find the new H

[tex]E_p = E_k[/tex]

[tex]mgH = mv_0^2/2[/tex]

[tex]H = \frac{v_0^2}{2g} = \frac{14.77^2}{2*9.81} = \frac{218.3}{19.62} = 11.245m[/tex]

the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.

When a projectile is launched at an angle, its vertical motion is influenced by gravity, while its horizontal motion is independent of gravity. In this case, the ball is initially launched at an angle of 58.0 degrees above the horizontal and reaches a maximum height of 8.0 m.

If the ball were thrown straight upward at the same initial speed (v0), its vertical motion would still be influenced by gravity. In both cases, the initial vertical speed (in the upward direction) is v0. The time it takes to reach the maximum height in both scenarios would also be the same because only the vertical component of velocity affects the vertical motion.

To find how high the ball would go if thrown straight upward, you can use the following kinematic equation:

h = (v0² * sin²(θ)) / (2 * g)

Where:

h is the maximum height (which we want to find).

v0 is the initial speed (given as v0).

θ is the launch angle (58.0 degrees).

g is the acceleration due to gravity (approximately 9.81 m/s²).

First, convert the angle from degrees to radians:

θ (in radians) = 58.0 degrees * (π radians / 180 degrees) ≈ 1.01 radians

Now, plug in the values and solve for h:

h = (v0² * sin²(1.01)) / (2 * 9.81 m/s²)

h ≈ (v0² * 0.454) / 19.62 m/s²

Now, if we consider that the initial speed v0 remains the same and throw the ball straight upward (θ = 90 degrees), the equation becomes:

h_straight_up = (v0² * sin²(π/2)) / (2 * 9.81 m/s²)

h_straight_up ≈ (v0² * 1) / 19.62 m/s²

Now, compare the two expressions for h and h_straight_up:

h / h_straight_up ≈ ((v0² * 0.454) / 19.62 m/s²) / ((v0² * 1) / 19.62 m/s²)

The "v0²" terms cancel out, and you get:

h / h_straight_up ≈ 0.454 / 1

h / h_straight_up ≈ 0.454

So, the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.

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Answer:

the phase difference is 1.26 radian

Solution:

As per the question:

Distance, d = 2 m

Distance from the other speaker, d' = 2.1 m

Frequency, f = 680 Hz

Speed of sound, v = 340 m/s

Now,

To calculate the phase difference, [tex]\Delta \phi[/tex]:

Path difference, [tex]\Delta d = d' - d = 2.1 - 2 = 0.1\ m[/tex]

For the wavelength:

[tex]f\lambda = v[/tex]

where

c = speed of light in vacuum

[tex]\lambda [/tex] = wavelength

Now,

[tex]680\times \lambda = 340[/tex]

[tex]\lambda = 0.5\ m[/tex]

Now,

Phase difference, [tex]\Delta phi = 2\pi \frac{\Delta d}{\lambda}[/tex]

[tex]\Delta phi = 2\pi \frac{0.1}{0.5} = 1.26\ rad[/tex]

Answer:

1590 N.

Explanation:

Force: This can be defined as the product of mass and the acceleration of a body. The S.I unit of Force is Newton (N).

The formula of force is given as

F = ma ........................ Equation 1

Where F = Force, m = mass of the astronaut, a = takeoff acceleration if the astronaut.

Given: m = 53.0 kg, a = 30 m/s²

Substitute into equation 1

F = 53(30)

F = 1590 N.

Hence the force exerted on the astronaut = 1590 N.

To calculate force, use the equation F=ma. Given an astronaut's mass of 53.0 kg and acceleration of 30.0 m/s² during a spaceship's takeoff, the exerted force totals around 1071 Newtons, subtracting the force of gravity.

To calculate the force exerted on the astronaut, you would use the equation F=ma, where F is force, m is mass and a is acceleration. Given that the mass (m) is 53.0 kg and the acceleration (a) is 30.0 m/s², the force (F) can be calculated as follows: F = ma = (53.0 kg)(30.0 m/s²) = 1590 Newtons.

This force is a result of combining gravity and the spaceship's upward propulsion (i.e. the astronaut's weight and the force of the spaceship accelerating). The force of gravity can be calculated simply by Fgravity = mg = (53.0 kg)(9.8 m/s²) = 519.4 Newtons. Therefore, the net force exerted by the spaceship on the astronaut during takeoff is F - Fgravity = 1590N - 519.4N = 1070.6 N, rounded off to 1071 N.

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Answer:

 R=19 N

Explanation:

Given that forces are taking opposite to each other.

Take

F₁ = 86 N ( Towards right ,take positive)

F₂ = - 67 N ( Towards left)

Given that the angle between above two forces is 180° so the resultant can be given as

Lets take the resultant = R N

R= F₁ + F₂

Now by putting the values

R= 86 N - 67 N

R=19 N

Therefore the resultant force on the block box will be 19 N.

Final answer:

The magnitude of the net force on the box is 19 N, calculated by subtracting the smaller force from the larger force since they are directed oppositely.

Explanation:

When two brothers push on a box from opposite directions, one with a force of 86 N and the other with a force of 67 N, the net force on the box is determined by subtracting the smaller force from the larger force because they are applied in opposite directions. Therefore, the magnitude of the net force on the box can be calculated as follows:

Net Force = Larger Force - Smaller Force
Net Force = 86 N - 67 N
Net Force = 19 N

Hence, the magnitude of the net force acting on the box is 19 N.

Answer:

-178.8 K

Explanation:

From Charles law,

V₁/T₁ = V₂/T₂.................... Equation 1

Where V₁ = Initial volume, T₁ = Initial Temperature, V₂ = Final volume, T₂ = Final Temperature.

Making T₂ the subject of the equation

T₂ = V₂T₁/V₁............... Equation 2

Given: V₂ = 1.00 L, V₁ = 2.5 L, T₁ = 298 K.

Substitute into equation 2

T₂ = 1.00(298)/2.5

T₂ = 119.2 K.

But,

Change in temperature = T₂ - T₁ = 119.2-298

Change in temperature = -178.8 K.

Hence the change in temperature = -178.8 K

The change in temperature of a system when its volume is reduced is calculated by utilizing Charles's law which states the volume of a gas is directly proportional to its absolute temperature under constant pressure.

To answer your query about the change in temperature when a system's volume is reduced, we would use Charles's law. Initially, the volume (V₁) is 2.50 L and the temperature (T₁) is 298 K. When the volume is reduced to 1.00 L (V₂), we're required to find the new temperature (T₂). According to Charles's law, V₁/T₁=V₂/T₂.

By manipulating this formula, we calculate the final temperature (T₂) as T₂=(V₂ * T₁)/V₁. Substituting the given values into this equation, we find T₂=(1.00 L * 298 K)/2.50 L. After performing the calculation, you will have the final temperature in Kelvin (K) when the volume is reduced to 1.00 L.

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Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

[tex]v_1=89km/h[/tex]

[tex]t_1=30min[/tex]

[tex]v_2=100km/h[/tex]

[tex]t_2=12min[/tex]

[tex]v_3=40km/h[/tex]

[tex]t_3=45 min[/tex]

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute=[tex]\frac{102}{60}=1.7 hour[/tex]

1 hour=60 minutes

Distance=[tex]speed\times time[/tex]

[tex]d_1=v_1\times t_1=80\times\frac{30}{60}=40km[/tex]

[tex]d_2=100\times \frac{12}{60}=20 km[/tex]

[tex]d_3=40\times \frac{45}{60}=30 km[/tex]

Total distance=[tex]d_1+d_2+d_3=40+20+30=90km[/tex]

Average speed=[tex]\frac{total\;speed}{total\;time}[/tex]

Using the formula

Average speed=[tex]\frac{90}{1.7}=52.9Km/h[/tex]

b.Total distance between the initial and final city lies along the route=90 km

Answer:

Electric field, E = 300.65 N/C

Explanation:

Given that,

Intensity of a beam of electromagnetic radiation, [tex]I=120\ W/m^2[/tex]

We need to find the maximum value of the electric field. The intensity of electromagnetic wave in terms of electric field is given by :

[tex]I=\dfrac{1}{2}\epsilon_oE^2c[/tex]

c is the speed of light

[tex]E=\sqrt{\dfrac{2I}{\epsilon_o c}}[/tex]

[tex]E=\sqrt{\dfrac{2\times 120}{8.85\times 10^{-12}\times 3\times 10^8}}[/tex]

E = 300.65 N/C

So, the maximum value of the electric field is 300.65 N/C. Hence, this is the required solution.

The maximum value of the electric field for the given beam of electromagnetic radiation.

the electric field of an electromagnetic radiation can be calculated by,

[tex]\bold {I = \dfrac 12 \epsilon_0 E^2C}[/tex]

[tex]\bold {E = \sqrt {\dfrac {2I}{\epsilon_0 C}}}[/tex]

Where,

I - intensity of the beam = 120 W/m2

C- speed of light = [tex]\bold { 3x10^8\ m/s}}[/tex]

E - electric field

Put the value of the formula

[tex]\bold {E = \sqrt {\dfrac {2\times 20 }{8.85x10^{-12} \times 3x10^8}}}\\\\\bold {E = 300.65\ N/C}[/tex]

Therefore, the maximum value of the electric field for the given beam of electromagnetic radiation.

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Answer:

changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

Explanation:

When the particle is removed from the wire, friction can be electrically charged, either with negative charges (extra electrons) or with positive charge by electron removal, in this case when the particle is between the condenser plates it experiences a force due to the electric field given by

          ΔV = E d

Where ΔV is the potential difference, d the distance between the plates and E the electric field.

In these cases we can use Newton's second law, where the acceleration is zero

                [tex]F_{e}[/tex] –W + B = 0

                 F_{e} = W –B

                 q E = mg - ρ_air g V

dodne B is the hydrostatic thrust

if  we know the density of the particular

                   ρ_particle = m / V

                   m = ρ_particle V

We replace

                 q E = g v (ρ_particle - ρ_air)

Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

To solve this exercise we will define the units of conversion between each of the variables given to facilitate the calculation in each section.

[tex]1 mile = 5280ft[/tex]

[tex]1h = 3600s[/tex]

[tex]1 ft = 30.48cm[/tex]

[tex]1 m = 100cm[/tex]

[tex]1kg = 1000g[/tex]

PART A ) For this part we have 60mph to ft/s, then

[tex]60mph = 60\frac{miles}{hour} (\frac{5280ft}{1mile})(\frac{1h}{3600s})[/tex]

[tex]60mph = 88ft/s[/tex]

PART B) Convert from [tex]ft/s^2[/tex] to [tex]m/s^2[/tex]

[tex]32ft/s^2 = 32 \frac{ft}{s^2} (\frac{30.48cm}{1ft})(\frac{1m}{100cm})[/tex]

[tex]32ft/s^2 = 9.7536m/s^2[/tex]

PART C) Convert [tex]g/cm^3[/tex] to [tex]kg/m^3[/tex]

[tex]1 g/cm^3 = 1\frac{g}{cm^3}(\frac{1kg}{1000g})(\frac{100cm}{1m})^3[/tex]

[tex]1 g/cm^3 = 1000kg/m^3[/tex]

Based on the conversion factors,  the conversions are as follows:

Conversions are used when converting between different units measuring the same quantity.

a. To convert 60 mph to ft/s:

60 mph = 60 * 5280/3600

60 mph = 88 ft/s

b. To convert 32 ft/s² to m/s

1 ft = 30.48 cm = 0.3048 m

32 ft/s² = 32 * 0.3034

32 ft/s² = 9.75 m/s²

c. To convert g/cm³ to kg/m³

1 g = 0.001 kg

1 cm³ = 0.000001 m³

1.0 g/cm3 = 1.0 * 0.001 kg/0.000001 m³

1 g/cm³ = 1000 Kg/m³

Therefore, the conversions are as follows:

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Answer:

If we increase the temperature in a reactor by 54 degrees Fahrenheit​ [54°F], the temperature will increase by 12.22 degrees Celsius [12.22 ⁰C]

Explanation:

To determine the number of degrees Celsius the temperature will be increased, we convert from Fahrenheit to Celsius.

Converting from Fahrenheit to degree Celsius

54°F -----> °C

54 = 1.8°C + 32

54-32 = 1.8°C

22 = 1.8°C

°C = 22/1.8

    = 12.22 °C

Thus, 54°F -----> 12.22 °C

Therefore, If we increase the temperature in a reactor by 54degrees Fahrenheit​ [54°F], the temperature will increase by 12.22 degrees Celsius [12.22 ⁰C]

12.22°C

The temperature increase in the reactor is 54°F

Now let's convert this to °C using the following relation;

(x − 32) × 5/9 = y         -------------------(i)

Where;

x is the value of the degree Fahrenheit = 54

y is the value of its corresponding degree Celsius.

Substitute x = 54 into equation (i)

(54 − 32) × 5/9 = y

(22) × 5/9 = y

Solve for y;

y = 22 x 5/9

y = 12.22°C

Therefore, the increase in temperature of the reactor in °C is 12.22

For the sinusoidal wave described by y = 0.25 sin (0.30x - 40t), the amplitude is 0.25 meters, angular frequency is 40 rad/s, the angular wave number is 0.30 rad/m and the wavelength is approximately 20.94 meters.

The given function for the sinusoidal wave is y = 0.25 sin (0.30x - 40t). We can extract the details of this wave from this function:

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Answer:

no additional information required.

Explanation:

given,

vertical height, h = 35 m

acceleration due to gravity, g = 9.8 m/s²

time taken by the bullet to fall in the water = ?

initial vertical velocity of the bullet = 0 m/s

using equation of motion for the time calculation

[tex]s = ut + \dfrac{1}{2}gt^2[/tex]

[tex]h= 0\times t + \dfrac{1}{2}\times g \times t^2[/tex]

[tex]t = \sqrt{\dfrac{2h}{g}}[/tex]

[tex]t= \sqrt{\dfrac{2\times 35}{9.8}}[/tex]

      t = 2.67 s

The time taken by the bullet to fall in the ocean is equal to 2.67 s.

Hence, there is no additional information required.

Answer:

0.00833542627085 Nm

Explanation:

[tex]\omega_f[/tex] = Final angular velocity = 700 rpm

[tex]\omega_i[/tex] = Initial angular velocity

[tex]\alpha[/tex] = Angular acceleration

m = Mass of stone = 1.3 kg

r = Radius = 8.5 cm

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-700\dfrac{2\pi}{60}}{41.3}\\\Rightarrow \alpha=-1.77491110372\ rad/s^2[/tex]

Torque is given by

[tex]\tau=-I\alpha\\\Rightarrow \tau=-\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=-\dfrac{1}{2}1.3\times 0.085^2\times -1.77491110372\\\Rightarrow \tau=0.00833542627085\ Nm[/tex]

The frictional torque is 0.00833542627085 Nm

Answer:

The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]

The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]

Explanation:

Given that,

Maximum shear strain = 0.005

We need to calculate the minimum shear strain

Using formula of maximum shear strain

[tex]\gamma_{max}=\dfrac{d_{2}}{2}\times\theta[/tex]

[tex]\theta=\dfrac{2\times\gamma_{max}}{d_{2}}[/tex]

Where, [tex]\gamma_{max}[/tex]=maximum shear strain

[tex]\theta[/tex]=angle of twist

[tex]d_{2}[/tex]= diameter

Put the value into the formula

[tex]\theta=\dfrac{2\times0.005}{3}[/tex]

[tex]\theta=0.00333\ rad[/tex]

[tex]\theta=3.33\times10^{-3}\ rad[/tex]

Now, Using formula of minimum shear strain

[tex]\gamma_{min}=\dfrac{d_{1}}{2}\times\theta[/tex]

Put the value into the formula

[tex]\gamma_{min}=\dfrac{2.5}{2}\times3.33\times10^{-3}[/tex]

[tex]\gamma_{min}=0.0041625[/tex]

[tex]\gamma_{min}=4.162\times10^{-3}[/tex]

We need to calculate the shear strain at the median line of the tube section

Using formula of shear strain at the median line

[tex]\gamma=\dfrac{d_{1}+d_{2}}{2}\times\theta[/tex]

Put the value into the formula

[tex]\gamma=\dfrac{2.5+3}{2}\times3.33\times10^{-3}[/tex]

[tex]\gamma=0.0091575[/tex]

[tex]\gamma=9.15\times10^{-3}[/tex]

Hence, The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]

The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]

Explanation of how to calculate the minimum shear strain and the shear strain at the median line in a circular tube subjected to torque.

Shear Strain Calculations:

Answer:

30 N

Explanation:

We are given that

Force,F =9 N

Charge,q=3 C

We have to find the exact magnitude of the force(in N) exerted by the system of charges on the 10.0 C.

We know that

[tex]E=\frac{F}{q}[/tex]

Using the formula

[tex]E=\frac{9}{3}=3 N/C[/tex]

Now, charge=10 C

The electric field remains same then

[tex]F=3\times 10=30 N[/tex]

Hence, the exact magnitude of the force exerted by the system of charges on the 10 C=30 N

The exact magnitude of the force exerted by the system of charges on the 10.0 C charge is 29.97 N.

The force exerted by a fixed system of charges on a charge is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, we are given that the force exerted on a 3.0 C charge is 9.0 N. When the 3.0 C charge is replaced with a 10.0 C charge, the force can be calculated using the proportionality of the charges. Since the charge is increased by a factor of (10.0 C)/(3.0 C) = 3.33, the force will also increase by the same factor. Therefore, the exact magnitude of the force exerted by the system of charges on the 10.0 C charge is 9.0 N * 3.33 = 29.97 N.

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Answer:

  α = 5 10⁻³ rad / s²

Explanation:

For this exercise we can use Newton's second law for rotational movement, where the force is electric

             τ = I α

Where the torque is

             τ = F x r = F r sin θ

Strength is

              F = q E

The moment of inertia of a small ball, which we approximate to a point is

             I = m r²

We replace

            2 (q E) r sin θ   = 2m r² α

The number 2 is because the two forces create the same torque

             α = q E sin θ / m r

Let's reduce the magnitudes to the SI system

           m = 1.0g = 1.0 10⁻³ kg

           L = 2.0 cm = 2.0 10⁻² m

           q = 10 nc = 10 10⁻⁹ C

           E = 1.0 10 N / C

           r = L / 2

           r = 1.0 10⁻² m

Let's calculate

           α = 10 10⁻⁹ 1.0 10 sin 30 / 1.0 10⁻³ 1.0 10⁻²

           α = 5 10⁻³ rad / s²

The initial angular acceleration of the system is 0.25 radians per second squared.

To find the initial angular acceleration of the system when the charged balls are released in the uniform electric field, we can use the concept of torque. The torque experienced by the system will cause it to rotate, resulting in angular acceleration.

The torque (τ) experienced by an electric dipole in an electric field is given by the formula:

τ = p * E * sin(θ)

Where:

- τ is the torque.

- p is the electric dipole moment, which is the product of the charge (q) and the separation (d) between the charges on the dipole: p = q * d.

- E is the electric field strength.

- θ is the angle between the electric field direction and the dipole moment direction.

In this case, we have:

- q1 = +10 nC (positive charge)

- q2 = -10 nC (negative charge)

- d = 2.0 cm = 0.02 m

- E = 1.0 * 10^3 N/C (1.0 * 10^3 N/C = 1.0 * 10^3 V/m, as 1 N/C = 1 V/m)

- θ = 30°

First, calculate the electric dipole moment (p):

p = q * d = (+10 * 10^-9 C) * (0.02 m) = 2 * 10^-10 C·m

Now, calculate the torque (τ):

τ = p * E * sin(θ) = (2 * 10^-10 C·m) * (1.0 * 10^3 V/m) * sin(30°)

To calculate sin(30°), we can use its exact value: sin(30°) = 1/2.

τ = (2 * 10^-10 C·m) * (1.0 * 10^3 V/m) * (1/2) = 1 * 10^-7 N·m

Now, we can find the moment of inertia (I) of the system. Since we have two balls rotating at a fixed distance (d), we can use the formula for a simple pendulum's moment of inertia:

I = 2 * m * d^2

Where:

- m is the mass of each ball (1.0 g = 0.001 kg)

- d is the separation (0.02 m)

I = 2 * (0.001 kg) * (0.02 m)^2 = 4 * 10^-7 kg·m^2

Now, we can calculate the initial angular acceleration (α) using the equation:

τ = I * α

α = τ / I = (1 * 10^-7 N·m) / (4 * 10^-7 kg·m^2) = 0.25 rad/s^2

So, the initial angular acceleration of the system is 0.25 radians per second squared.

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The question involves calculations using the principles of projectile motion and 2D kinematics to understand the trajectory of a baseball pitch. Values include horizontal and vertical velocity changes over time, the exact speed of the baseball and the angle at which baseball hits the ground.

The question involves the physics of projectile motion, specifically applied to the trajectory of a pitch in baseball. The parameters of this problem can be solved using the kinematic equations and principles that govern 2D motion.

(1) To solve part (a), one must understand that for such problems involving gravity, the x (horizontal) component of the object's velocity remains constant throughout the motion, in this case 142 ft/sec. This fact is obtained from the equation x(t) = v_xt, where v_x is the constant x-component (horizontal) of velocity.

(2) For part (b), similarly, the x-component of the velocity remains same at any point in trajectory, including when it passes over home plate, so it is again 142 ft/sec.

(3) For part (c), the derivative of y(t) -16t^2 + 5t + 5 needs to be found to get the y-component (vertical) of velocity which is -32t + 5. This is because as the object moves under the effect of gravity, vertical velocity changes with time.

(4) For part (d), the magnitude of the speed at any time can be found by taking the square root of the sum of squares of x and y-components of velocity. (5) Part (e), seeks the time when the baseball hits the ground. The equation of vertical motion -16t^2 + 5t + 5 = 0 should be used, as the height of the object from ground is 0 when it hits the ground. Solving for t would give that time.

End of the problem involves understanding some trigonometric relationships to calculate the angle (in radians).

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Answer:

17.23 m/s²

Explanation:

Applying the equation of motion,

Δs = ut + 1/2at²..................... Equation 1

Where Δs  = change in height of the rocket, u = initial velocity of the rocket, a = acceleration of the rocket, t = time

making a the subject of the equation,

a = 2(Δs-ut)/t²..................... Equation 2

Given: Δs = (56-4) m = 52 m, u = 11.0 m/s, t = 1.90 s.

Substitute into equation 2

a = 2[52-(11×1.9)]/1.9²

a = 2(52-20.9)/1.9²

a = 2(31.1)/3.61

a = 62.2/3.61

a = 17.23 m/s².

Thus the acceleration = 17.23 m/s²

Final answer:

To find the magnitude of acceleration, we used the kinematic equation s = ut + ½at². We determined the displacement due to the initial velocity and the time interval, and then solved for acceleration to find that the rocket's magnitude of acceleration is 17.2 m/s².

Explanation:

To calculate the magnitude of its acceleration, we can use the kinematic equations for uniformly accelerated motion along with the information provided about the rocket's velocity and positions at different times. We have two positions, 4.00 m and 56.0 m, and a time interval of 1.90 s. The rocket's velocity at the lower height is 11.0 m/s.

Step 1: Use the second kinematic equation

We will use the kinematic equation:
s = ut + ½at², where 's' is the displacement, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.

Step 2: Set up the equation with the known values

Let's calculate the displacement (Δs): Δs = 56.0 m - 4.00 m = 52.0 m
Now put the known values into the equation: 52.0 m = (11.0 m/s)(1.90 s) + ½a(1.90 s)²

Step 3: Solve for acceleration 'a'

First, we calculate the distance traveled due to the initial velocity: (11.0 m/s)(1.90 s) = 20.9 m
Subtract this value from the total displacement to find the displacement caused by acceleration: 52.0 m - 20.9 m = 31.1 m
Rewrite the equation with this new information: 31.1 m = ½a(1.90 s)²
Now solve for 'a': a = (2 × 31.1 m) / (1.90 s)² = 17.2 m/s²

The magnitude of acceleration of the rocket is therefore 17.2 m/s².

Answer:

(a) 2.154×10⁻¹¹ A.

(b)  98300000.

Explanation:

(a)

Using Ohm's law,

V = IR ........................ Equation 1

Where V = Potential difference between the walls, I = current, R = Resistance between the walls.

Make I the subject of the equation

I = V/R...................... Equation 2

Given: V = 84 mV, = 0.084 V, R = 3.9×10⁹ Ω.

Substitute into equation 2

I = 0.084/(3.9×10⁹)

I = 2.154×10⁻¹¹ A.

(b)

I = q/t

q = It .................... Equation 1

Where q = quantity of electric charge, t = time.

Given: I = 2.154×10⁻¹¹ A, t = 0.73 s.

q = 2.154×10⁻¹¹×0.73

q = 1.572×10⁻¹¹ C.

The charge on an electron e = 1.6×10⁻¹⁹ C

n = q/e

where n = number of ions.

n = 1.572×10⁻¹¹/1.6×10⁻¹⁹

n = 9.83×10⁷

n = 98300000.

Answer

given,

Speed of the Aircraft,v = 175 mi/h

1 mi/h = 0.44704 m/s

175 mi/h = 78.232 m/s

time, t = 2.5 s

a) average acceleration = ?

[tex]a= \dfrac{v - u}{t}[/tex]

[tex]a= \dfrac{78.232 - 0}{2.5}[/tex]

a = 31.29 m/s²

b) Distance traveled by the Pane

using equation of motion

v² = u² + 2 a s

78.232² = 0² + 2 x 31.29 x s

s = 97.79 m

Distance moved by the plane is equal to 97.79 m

Answer:

The acceleration of the laundry cart is 13.3 m/s²

Explanation:

Hi there!

According to Newton's second law, the sum of all forces acting on an object in a given direction is equal to the mass of the object times its acceleration in that direction:

∑F = m · a

Where:

∑F = net force acting on the cart.

m = mass of the cart.

a = acceleration.

Then, solving this equation for the acceleration:

∑F / m = a

60.0 N / 4.50 kg = a

a = 13.3 m/s²

The acceleration of the laundry cart is 13.3 m/s²

Using the formula of acceleration a = F/m, where F is the net external force and m is the mass, substituting the given values yields an acceleration of 13.33 m/s².

According to Newton's second law, the acceleration of an object as produced by a net force is directly proportional to the net force, in the same direction as the net force, and inversely proportional to the mass of the object. This can be expressed with the formula a = F/m, where F is the net external force and m is the object's mass.

Given the mass m of the laundry cart is 4.5 kg and the net external force F on the cart is 60.0 N, we can substitute these values into the formula.

Therefore, the acceleration a = F/m = 60.0 N / 4.50 kg = 13.33 m/s².

So, the magnitude of the cart's acceleration is 13.33 m/s².

Learn more about Newton's Second Law here:

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Using kinematics equations, we can calculate the time for a rock to fall from the cliff on Earth. Without the value of acceleration due to gravity on the extrasolar planet, we can't quantify the time difference. A difference would exist if the gravitational accelerations of the two planets differ.

This question pertains to kinematics and physical laws regarding free fall. On earth, when the astronaut throws a rock straight up with a certain velocity, initially, gravity will slow down the rock and finally make it fall back toward the ground. This action is governed by the equation of motion: h = vit + 1/2gt² where h is the height, vi the initial velocity, t the time and g the acceleration due to gravity on earth (-9.81 m/s²).

In the extrasolar planet scenario, considering it is not mentioned, we could assume that the acceleration due to gravity might be similar to earth's. However, without specific details about the gravity, we can still calculate the time it takes for the rock to land from the 50m high cliff using the motion equation for a free-falling body. We can conclude that if the gravity in the two settings is different, then there would be a time difference. We can’t quantify the time difference without knowing the value of the gravity on the extrasolar planet.

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To solve this problem we will apply the concept of balance of Forces in the body. For such an effect the centripetal force must be equivalent to the electrostatic force of the body, therefore

[tex]F_c = F_e[/tex]

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

Here

m = Mass of electron

r = Distance between them

k = Coulomb's constant

[tex]q_p[/tex] = Charge of proton

[tex]q_e[/tex] = Charge of electron

v = Velocity

Rearranging to find the velocity we have that,

[tex]v^2 = \frac{kq_pq_e}{mr}[/tex]

[tex]v = \sqrt{\frac{kq_pq_e}{mr}}[/tex]

Replacing,

[tex]v = \sqrt{\frac{(9*10^9)(1.6*10^{-19})(1.6*10^{-19})}{(9.1*10^{-31})(5.29*10^{-11})}}[/tex]

[tex]v = 2.19*10^6m/s[/tex]

Therefore the speed of the electron is [tex]2.19*10^6m/s[/tex]

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

[tex]K.E=\frac{3}{2}KT[/tex]

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

[tex]T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K[/tex]

converting to degree we have [tex]131^{0}C[/tex]

Answer:

a) Earth

Explanation:

When constructing an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be Earth.

From the given options the earth is the closest planet to the center of the field. The center of the field can be considered as the sun. In reality we have mercury as the closest planet in our  solar system.

Answer:

Angular velocity, [tex]\omega=3747.33\ rev/min[/tex]

Explanation:

In this case, we need to find the angular speed needed for a centrifuge to produce an acceleration of 759 times the gravitational acceleration.

Radius of the circular path, r = 4.83 cm

The acceleration acting on the particle in circular path is given by :

[tex]a=r\omega^2[/tex]

[tex]\omega[/tex] is the angular speed in rad/s

[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]

[tex]\omega=\sqrt{\dfrac{759\times 9.8}{4.83\times 10^{-2}}}[/tex]

[tex]\omega=392.42\ rad/s[/tex]

or

[tex]\omega=3747.33\ rev/min[/tex]

So, there are 3747.33 revolutions per minute that is needed. Hence, this is the required solution.

Answer:

The other frequency is either 253Hz (the altered tuning frequency vibrates lower) or 259Hz(if the altered tuning frequency vibrates higher):

Explanation:

We have two frequencies: the tuning fork frequency  and  the altered tuning fork frequency  

F1 -----tuning fork frequency

F2-----Altered tuning fork frequency

Fbeat=3

F1=256Hz

To find the alterd tuning fork frequency:

The beat frequency of any two waves is:

Fbeat =|f1 – f2|

i.e  we have two choices

for  frequency tied to length increase (wavelength increases as length increases, therefore frequency decrease (the altered tuning frequency vibrates lower):   Fbeat = f2-f1

       F2=F1-Fbeat

     F2= 256-3

         =253Hz

For  frequency tied to length decrease wavelength increases as length increases, therefore frequency also increases(the altered tuning frequency vibrates higher) :    Fbeat = -(f2-f1)

F2= Fbeat+F1

   = 256+3

   = 259Hz

Note that a wavelength increase means a decrease of frequency because v = fλ

Answer:

Part a: The schematic diagram is attached.

Part b: The pressure at the end is 1 bar.

Part c: Weight of 0.5kmol of water is 88.2 N.

Part d: The specific volume is 0.001 m^3/kg

Explanation:

The schematic is given in the diagram attached.

Pressure is given using the ideal gas equation as

Here

                      [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{1\times V_1}{300}=\frac{P_2\times 2V_1}{600}\\P_2=\frac{600}{600}\\P_2=1 bar[/tex]

So the pressure at the end is 1 bar.

Mass of 0.5kmol is given as follows

                              [tex]Mass=n_{moles} \times Molar \, Mass\\Mass=0.5 \times 10^3 \times 18 \times 10^{-3}\\Mass=9.0 kg[/tex]

Weight is given as

                           [tex]W=mxg\\W=9 \times 9.8\\W=88.2 \, N[/tex]

So weight of 0.5kmol of water is 88.2 N.

Specific volume is given as

                        [tex]v=\frac{Volume}{Mass}\\v=\frac{0.009}{9}\\v=0.001 m^3/kg[/tex]

So the specific volume is 0.001 m^3/kg

A. Draw a schematic of the system with a boundary around the piston-cylinder device. B. The final pressure can be determined using the ideal gas law equation. C. The weight of H2O can be calculated using the relationship between moles, molecular mass, and mass. D. The specific volume of H2O can be determined by dividing the volume by the mass.

A. To study the air in the piston-cylinder device as a closed system, we consider the device itself as the system and draw a boundary around it, including the air inside and excluding the surroundings. The schematic would show a cylindrical container with a piston separating the initial and final air volumes.

B. To determine the final pressure of the air, we can use the ideal gas law. The equation is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. Since the volume doubles and the temperature increases to 600 K, we can set up the equation (1 bar)(2V) = n(R)(600 K), and solve for the final pressure.

C. To determine the weight of H2O in the piston-cylinder device, we use the relationship between number of moles, molecular mass, and mass. The weight of H2O in N can be calculated as (0.5 kmol)(molecular mass of H2O)(Acceleration due to gravity).

D. The specific volume of H2O can be determined by dividing the volume (0.009 m3) by the mass of H2O (which we can calculate from the number of moles and molecular mass).

Answer:

W = 1.44 10⁻⁷ J

Explanation:

The expression for the job is

          W = ∫ F. dx

Where the point is the scalar product in this case the direction of the meteor and the depth is parallel, whereby the scalar product is reduced to the ordinary product

          W = 630 ∫ x³ dx

          W = 630 x⁴ / 4

Let's evaluate between the lower limit x = 0, w = 0 to the upper limite the point at x = 5.5 10⁻³ m

            W = 157.5 ((5.5 10⁻³)⁴ -0)

            W = 1.44 10⁻⁷ J