a ball is thrown at an angle of 45° to the ground. if the ball lands 89 m away, what was the initial speed of the ball? (round your answer to the nearest whole number. use g ≈ 9.8 m/s2.) v0 = m/s

Answers

Answer 1

The initial speed of the ball was approximately 199 m/s.

Explain the projectile motion?

Projectile motion is the motion of an object that is launched into the air and then moves under the influence of gravity alone.

When a ball is thrown at an angle to the ground, its motion can be divided into two components: a horizontal component and a vertical component. The horizontal component is constant and equal to the initial velocity multiplied by the cosine of the angle of projection, while the vertical component is affected by gravity and changes over time.

In this problem, the ball is thrown at an angle of 45° to the ground, which means that the horizontal and vertical components of its initial velocity are equal. Therefore, we can write:

[tex]vx = v0 cos(45°)[/tex]

[tex]vy = v0 sin(45°)[/tex]

where [tex]vx[/tex] is the horizontal component of the initial velocity, [tex]vy[/tex] is the vertical component of the initial velocity, and v0 is the magnitude of the initial velocity.

Now, we can use the fact that the ball lands 89 m away to find the time it takes for the ball to travel that distance. Since there is no air resistance, the time of flight of the ball is equal to twice the time it takes for the ball to reach its maximum height. This can be found using the following kinematic equation:

[tex]y = vy*t - (1/2)gt^2[/tex]

where y is the vertical displacement of the ball, t is the time elapsed since the ball was thrown, and g is the acceleration due to gravity.

At the maximum height, the vertical displacement of the ball is given by:

[tex]ymax = (v0 sin(45°))^2 / (2g)[/tex]

The time it takes for the ball to reach its maximum height can be found by setting y = ymax and solving for t:

[tex]t = vy / g = v0 sin(45°) / g[/tex]

The time of flight of the ball is then:

[tex]T = 2t = 2v0 sin(45°) / g[/tex]

During the time of flight, the horizontal displacement of the ball is given by:

[tex]x = vx*T = v0 cos(45°) * 2v0 sin(45°) / g = 2v0^2 / g[/tex]

Setting x = 89 m and solving for v0, we get:

[tex]v0 = √(89*g/2) / sin(45°)[/tex] [tex]= 199m/s[/tex]

Therefore, the initial speed of the ball was approximately 199 m/s.

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Related Questions

for a particular process at 300 k, δg δ g = -10.0 kj and δh δ h = -7.0 kj. if the process is carried out reversibly, the amount of useful work (in kj) that .

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The amount of useful work that can be obtained for this particular process at 300 K is -3.0 kJ.

For a reversible process, the amount of useful work (in kJ) that can be obtained is given by the equation:

w = -δg = δh - Tδs

where δs is the change in entropy of the system.

Since the process is carried out reversibly, δs can be calculated using the equation:

δs = δqrev / T

where δqrev is the heat absorbed by the system during a reversible process.

Since δh = -7.0 kJ and δg = -10.0 kJ, we know that the process is exothermic (δh < 0) and spontaneous (δg < 0). Therefore, δqrev must be negative, indicating that heat is released from the system.

We can calculate δqrev using the equation:

δqrev = -Tδs = -T(δh / T) = -δh = 7.0 kJ

Substituting this value into the equation for work, we get:

w = -δg = δh - Tδs = -10.0 kJ - (-7.0 kJ) = -3.0 kJ

Therefore, the amount of useful work for this particular process at 300 K is -10.0 kJ.

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A closed curve encircles several conductors. The line integral around this curve is ∮B⃗ ⋅dl⃗ = 4.25×10^-4 T⋅m .
A) What is the net current in the conductors?
B) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?

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a) the net current in the conductors is 3.38 A. b) the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

To solve this problem, we can use Ampere's Law, which relates the line integral of the magnetic field around a closed loop to the net current passing through the loop.

A) The equation for Ampere's Law is: ∮B⃗ ⋅dl⃗ = μ0I, where μ0 is the permeability of free space and I is the net current passing through the loop. Solving for I, we get:

I = ∮B⃗ ⋅dl⃗ / μ0

Substituting the given values, we get:

I = (4.25×10^-4 T⋅m) / (4π×10^-7 T⋅m/A)

I = 3.38 A

Therefore, the net current in the conductors is 3.38 A.

B) If we integrate around the curve in the opposite direction, the value of the line integral will be negative, since the direction of the magnetic field will be opposite. Specifically, we can use the fact that reversing the direction of the line integral is equivalent to reversing the direction of the loop, which changes the sign of the enclosed current.

Therefore, the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

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An inductor is connected to a 12 khz oscillator that produces an rms voltage of 8.0 v. The peak current is 80 ma. What is the value of the inductance L?

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The value of the inductance L is 4.77 mH.

To find the value of the inductance L, first, we need to determine the reactance (X_L) of the inductor using the given rms voltage (8.0 V) and peak current (80 mA). The relationship between voltage, current, and reactance is given by Ohm's law for inductive circuits: V_rms = I_peak × X_L.

1. Convert peak current to rms current: I_rms = I_peak / √2 = 80 mA / √2 ≈ 56.57 mA
2. Calculate reactance: X_L = V_rms / I_rms = 8.0 V / 56.57 mA ≈ 141.42 Ω
3. Find inductance using the formula X_L = 2π × frequency × L:
L = X_L / (2π × frequency) = 141.42 Ω / (2π × 12 kHz) ≈ 4.77 mH

The value of the inductance L in the given scenario is 4.77 millihenries (mH).

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You will need to design a cascode amplifier, which should be built and tested to meet the following requirements:1. Magnitude of the voltage gain = 12*SQRT(Z+35):± 10%, where Z is the sum of the last 3 digits of yourstudent number.2. The load resistance RL = 6*(Z+40)2 Ω, rounded up to the nearest standard value stocked in the lab, i.e.,decade multiples of 1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, and 8.2 kΩ. As an example, if yourRL is 67.4 kΩ, you will round this to 68 kΩ.3. The high frequency cutoff fH is to be maximized. It must exceed 1 MHz.4. The output voltage should be able to get to 2 V peak-peak without appreciable distortion[1]. To ensurethis, the AC base-emitter voltage must be kept under 10 mV peak-peak for such an output.5. No DC current may flow in RL and no DC current may flow into or out of the signal generator.6. The low frequency fL must be less than 200 Hz.7. The input and output impedances are left to the discretion of the designer, but their magnitudes at 1kHz are to be determined by calculation and then measured.8. Total circuit power is not to exceed 50 mW.9. The transistors are all to be 2N3904.10. Collector currents in the transistors are to be 1.0 mA ±10%.11. Power-supply voltages are to be limited to +5 volts and/or +15 volts and/or -15 volts.12. No adjustable components, e.g. a trimmer potentiometer, will be allowed.13. Available capacitors are limited to: 1 x 100 µF, 1 x 33 µF, 1 x 10 µF, 2 x 1 µF, 1 x 0.1 µF [2].14. The choice of all other components is left up to the designer.

Answers

To design a cascode amplifier meeting the specified requirements, follow these steps:

1. Calculate the voltage gain (Av) using the formula Av = 12 * SQRT(Z + 35), where Z is the sum of the last 3 digits of your student number.
2. Determine the load resistance (RL) using the formula RL = 6 * (Z + 40)² Ω, and round up to the nearest standard value stocked in the lab.
3. Use a high-pass filter at the input to achieve the low frequency cutoff (fL) of less than 200 Hz.
4. Design the cascode amplifier using 2N3904 transistors with collector currents set to 1.0 mA ±10%. The power supply voltages should be limited to +5V, +15V, or -15V.
5. Maximize the high frequency cutoff (fH) to exceed 1 MHz by carefully selecting component values and minimizing parasitic capacitances.
6. Ensure the output voltage can reach 2 V peak-peak without distortion by keeping the AC base-emitter voltage below 10 mV peak-peak.
7. Prevent DC current from flowing in RL and the signal generator by using coupling capacitors.
8. Calculate and measure the input and output impedances at 1 kHz.
9. Limit the total circuit power to 50 mW.
10. Use the available capacitors (1 x 100 µF, 1 x 33 µF, 1 x 10 µF, 2 x 1 µF, 1 x 0.1 µF) in the design.
11. Choose all other components as needed to achieve the desired performance while adhering to the constraints.

By following these steps, you will design a cascode amplifier that meets the given requirements. Remember, no adjustable components are allowed, and all transistors must be 2N3904.

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what is the sensitivity (in µa) of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 1.00 mω resistance on its 30.5 v scale?

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The sensitivity of the galvanometer inside the voltmeter is 30,500,000 µA.

Hi! To find the sensitivity (in µA) of the galvanometer inside a voltmeter with a 1.00 mΩ resistance on its 30.5 V scale, you can follow these steps:

1. First, note the full-scale voltage, V = 30.5 V, and the internal resistance of the voltmeter, R = 1.00 mΩ.
2. Use Ohm's law to calculate the current for full-scale deflection, I = V/R.
3. Convert the calculated current to microamperes (µA).

Now, let's calculate the sensitivity of the galvanometer:

1. V = 30.5 V, R = 1.00 mΩ = 0.001 Ω (since 1 mΩ = 0.001 Ω).
2. I = V/R = 30.5 V / 0.001 Ω = 30500 A.
3. Convert the current to µA: 1 A = 1,000,000 µA, so 30500 A = 30,500,000 µA.

So, the sensitivity of the galvanometer inside the voltmeter is 30,500,000 µA.

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the current is from left to right in the conductor showmn. the magnetic field is onto the page and point is at higher potential than point t. the charge carriers are
a. positive b. negative c. neutral d. absent e. moving near the speed of light

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The charge carriers in the conductor are likely to be negative, and they are not absent, but their speed is not near the speed of light due to the effects of collisions in the conductor.

Based on the given information, we know that the conductor is experiencing a magnetic force due to the magnetic field pointing onto the page. Additionally, we know that the point labeled as "point" is at a higher potential than the point labeled as "t".

The direction of the current flowing from left to right in the conductor is indicative of the direction in which the charge carriers are moving. Given this information, we can determine that the charge carriers in the conductor must be negative because electrons are negatively charged and move opposite to the direction of conventional current flow.

Furthermore, the fact that the charge carriers are moving in a conductor does not necessarily imply that they are moving near the speed of light. The speed at which electrons move in a conductor is known as the drift velocity and is typically much slower than the speed of light due to collisions with other particles in the conductor.

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In a parallel circuit, the current through any resistor is directly proportional to the value of that resistor. True or False.

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False. In a parallel circuit, the current through each resistor is not directly proportional to the value of that resistor. Instead, the current through each resistor is inversely proportional to the value of the resistor. This means that as the resistance of a resistor increases, the current through it decreases, and vice versa.

In a parallel circuit, the voltage across each resistor is the same, but the current flowing through each resistor can be different. This is because each resistor provides a different path for the flow of electricity. The total current flowing into the parallel circuit is divided among the different resistors according to their individual resistance values.

In summary, in a parallel circuit, the current through each resistor is inversely proportional to its value. This is because the total current flowing into the parallel circuit is divided among the different resistors according to their individual resistance values, and the voltage across each resistor is the same.

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show that the transition matrix is regular and find its steady-state vector.

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To show that a transition matrix is regular by raising the matrix to different powers until you find a power with all positive elements and its steady-state vector is equation πP = π while ensuring the elements of π sum to 1.

A transition matrix is regular if some power of the matrix has only positive elements, a steady-state vector is a probability vector that remains unchanged after being multiplied by the transition matrix. To demonstrate that the transition matrix is regular, raise the matrix to different powers until you find a power with all positive elements. If such a power exists, the matrix is considered regular.

Next, to find the steady-state vector, solve the following equation: πP = π, where π is the steady-state vector, and P is the transition matrix. Additionally, ensure the elements of π sum to 1, representing the total probability, you can solve this system of linear equations using methods like Gaussian elimination, matrix inversion, or iterative techniques. In summary, to show that a transition matrix is regular, find a power of the matrix with all positive elements, then, to find the steady-state vector, solve the equation πP = π while ensuring the elements of π sum to 1.

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A typical power for a laser used in physics labs is 0.75 mW. This laser would produce a beam that is about 1 mm in diameter I found the average intensity: to be 955 W/m2 I found the average energy density of this beam: 3.183*10^-6 J/m^3 I need help with this question. Lets say the laser beam is reflected completely off a mirror. What is the maximum force the beam can exert on the mirror? I did 955 x 2x (4pie x 10^-7)/(3*10^8) but says my answer is wrong, is this the right equation???

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The maximum force of a typical power for a laser used in physics labs is 0.75 mW and would produce a beam that is about 1 mm in diameter can exert on the mirror is 5 x 10⁻⁹ N.

In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

To find the maximum force the laser beam can exert on the mirror when it is reflected completely, you should use the following formula:

Force (F) = (2 × Power (P)) / Speed of light (c)

where Power (P) = 0.75 mW (convert to Watts: 0.75 x 10⁻³ W), and Speed of light (c) = 3 x 10⁸ m/s.

F = (2 × (0.75 x 10⁻³ W)) / (3 x 10⁸ m/s)

Thus, the maximum force of the beam can exert on the mirror is approximately 5 x 10⁻⁹ N.

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Find the value of “F2”

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The reaction force exerted by m₁ is 118.4 N.

Mass of the upper block, m₁ = 8 kg

Mass of the lower block, m₂ = 15 kg

Acceleration, a = 5 m/s₂

Normal reaction is a force that applies perpendicularly to two surfaces that are in contact. It represents the force that is holding the two surfaces together.

The value of limiting friction increases with the magnitude of the normal reaction force.

The force exerted by m₁ is,

F₁ = m₁(g + a)

F₁ = 8(9.8 + 5)

F₁ = 118.4 N

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Your question was incomplete. Attaching the image file here

Johnny, of mass 65 kg, and Lucy, of mass 45 kg, are facing each other on roller blades. The coefficient of kinetic friction between the roller blades and concrete surface is 0.20. When Johnny pushes Lucy from rest he applies a force for 1.0 s. Lucy then slows down to a stop in another 8.0 s. Calculate:
a. The applied force exerted by Johnny on Lucy.
b. How long it takes Johnny to come to rest.

I tried calculated the force exerted but I would need acceleration which I don't have...any tips on how to solve this one??? help is appreciated!!

Answers

Answer:

John applied a force of approximately [tex]795\; {\rm N}[/tex] (on average, rounded) on Lucy.

John slows down to a stop after approximately another [tex]5.37\; {\rm s}[/tex].

(Assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)

Explanation:

Assuming that the surface is level. The normal force on Johnny will be equal to the weight of Johnny: [tex]N(\text{John}) = m(\text{John})\, g[/tex]. Similarly, the normal force on Lucy will be equal to weight [tex]N(\text{Lucy}) = m(\text{Lucy})\, g[/tex].

Multiply normal force by the coefficient of kinetic friction to find the friction on each person:

[tex]f(\text{John}) = \mu_{k}\, N(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex].

[tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex].

Again, because the surface is level, the net force on each person after the first [tex]1.0\; {\rm s}[/tex] will be equal to the friction. Divide that the net force on each person by the mass of that person to find acceleration:
[tex]\displaystyle a(\text{John}) = \frac{\mu_{k}\, m(\text{John})\, g}{m(\text{John})} = \mu_{k}\, g[/tex].

[tex]\displaystyle a(\text{Lucy}) = \frac{\mu_{k}\, m(\text{Lucy})\, g}{m(\text{Lucy})} = \mu_{k}\, g[/tex].

(Note that the magnitude of acceleration is independent of mass and is the same for both John and Lucy.)

[tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex].

In other words, after the first [tex]1\; {\rm s}[/tex], both John and Lucy will slow down at a rate of [tex]1.962\; {\rm m\cdot s^{-2}}[/tex].

To find the speed of Lucy immediately after the first [tex]1.0\: {\rm s}[/tex], multiply this acceleration by the time [tex]t = 8.0\; {\rm s}[/tex] it took for Lucy to slow down to [tex]0\; {\rm m\cdot s^{-1}}[/tex]:

[tex]\begin{aligned}& (8.0\; {\rm s})\, (1.962\; {\rm m\cdot s^{-2}}) \\ =\; & (8.0)\, (1.962)\; {\rm m\cdot s^{-1}} \\ =\; & 15.696\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Thus, in the first [tex]1.0\; {\rm s}[/tex], Lucy accelerated (from [tex]0\; {\rm m\cdot s^{-1}}[/tex]) to [tex]15.696\; {\rm m\cdot s^{-1}}[/tex].

The average acceleration of Lucy in the first [tex]1.0\; {\rm s}[/tex] would be [tex](15.696) / (1) = 15.696\; {\rm m\cdot s^{-2}}[/tex]. Multiply this average acceleration by the mass of Lucy to find the average net force on Lucy during that [tex]1.0\; {\rm s}[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{Lucy}) &= m(\text{Lucy})\, a \\ &= (45)\, (15.696)\; {\rm N} \\ &= 706.320\; {\rm N}\end{aligned}[/tex].

This net force on Lucy during that [tex]1.0\; {\rm s}[/tex] is the combined result of both the push from Johnny and friction:

[tex]F_{\text{net}}(\text{Lucy}) = F(\text{push}) - f(\text{Lucy})[/tex].

Since [tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex]:

[tex]\begin{aligned}F(\text{push}) &= F_{\text{net}}(\text{Lucy}) + f(\text{Lucy}) \\ &= F_{\text{net}}(\text{Lucy}) + \mu_{k}\, m(\text{Lucy})\, g \\ &= (706.320) \; {\rm N}+ (0.2)\, (45)\, (9.81)\; {\rm N} \\ &= 706.320\; {\rm N} + 88.290\; {\rm N} \\ &=794.610\; {\rm N}\end{aligned}[/tex].

In other words, Johnny would have applied a force of [tex]794.610\; {\rm N}[/tex] on Lucy.

By Newton's Laws of Motion, when Johnny exerts this force on Lucy in that [tex]1.0\; {\rm s}[/tex], Lucy would exert a reaction force on Johnny of the same magnitude: [tex]794.610\; {\rm N}[/tex].

Similar to Lucy, the net force on Johnny during that [tex]1.0\; {\rm s}[/tex] will be the combined effect of the push [tex]F(\text{push})[/tex] and friction [tex]f(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{John}) &= F(\text{push}) - f(\text{John}) \\ &= F(\text{push}) - \mu_{k}\, m(\text{John})\, g\\ &= 794.610\; {\rm N} - (0.2)\, (65)\, (9.81)\; {\rm N} \\ &= 667.080\; {\rm N}\end{aligned}[/tex].

Divide net force by mass to find acceleration:

[tex]\begin{aligned}\frac{667.080\; {\rm N}}{65\; {\rm kg}} \approx 10.2628\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

In other words, Johnny accelerated at a rate of approximately [tex]10.5406\; {\rm m\cdot s^{-2}}[/tex] during that [tex]1.0\; {\rm s}[/tex]. Assuming that Johnny was initially not moving, the velocity of Johnny right after that [tex]1.0\; {\rm s}\![/tex] would be:

[tex](0\; {\rm m\cdot s^{-1}}) + (10.2628\; {\rm m\cdot s^{-2}})\, (1.0\; {\rm s}) = 10.2628\; {\rm m\cdot s^{-1}}[/tex].

After the first [tex]1.0\; {\rm s}[/tex], the acceleration of both John and Lucy (as a result of friction) would both be equal to [tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex]. Divide initial velocity of Johnny by this acceleration to find the time it took for Johnny to slow down to a stop:

[tex]\displaystyle \frac{10.2628\; {\rm {m\cdot s^{-1}}}}{1.962\; {\rm m\cdot s^{-2}}} \approx 5.23\; {\rm s}[/tex].