Light bulb 1 operates with a filament temperature of 2800 K, whereas light bulb 2 has a filament temperature of 1700 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs. A1/A2

Answer:39.2 (B)

Explanation:

Answer:

the answer is D.

Explanation: Mark me brainlest

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  [tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

On putting estimate values, we get

⇒  [tex]\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}[/tex]

⇒  [tex]\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}[/tex]

⇒  [tex]v=1.94 \ cm[/tex]

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

[tex]m=\frac{v}{u}[/tex] and [tex]m=\frac{h_{1}}{h_{0}}[/tex]

⇒  [tex]\frac{v}{u} =\frac{h_{1}}{h_{0}}[/tex]

⇒  [tex]\frac{1.94}{25}=\frac{{h_{1}}}{15}[/tex]

⇒  [tex]{h_{1}}=1.16 \ mm[/tex]

Answer:

c

Explanation:

Answer:

The circuit is in series connection,

the same current will flow through i.e

I1 = I2 = I3 = 2A.

Explanation:

We'll begin by calculating the total resistance in the circuit. This is shown below:

R1 = 4Ω

R2 = 3Ω

RT =..?

Total resistance in series can be obtained as follow:

RT = R1 + R2

RT = 4 + 3

RT = 7Ω

Next we shall determine the total current flowing in the circuit:

Voltage (V) = 14V

Resistance (R) = 7Ω

Current (I) =..?

V = IR

14 = I x 7

Divide both side by 7

I = 14/7

I = 2A.

Since the circuit is in series connection,

the same current will flow through i.e

I1 = I2 = I3 = 2A.

Answer:

On the Moon, the jump height is about 6 times higher than on Earth.

Explanation:

The maximum height reached by an object is given by the formula as follows :

[tex]H=\dfrac{v^2}{2g}[/tex]

v is the speed of object

g is acceleration due to gravity on Earth's surface

If g' is acceleration on surface of Moon, g' = g/6

Let H' is the height reached by object on Moon, then,

[tex]H'=\dfrac{v^2}{2g'}\\\\H'=\dfrac{v^2}{2\times \dfrac{g}{6}}\\\\H'=6\times \dfrac{v^2}{2g}\\\\H'=6H[/tex]

On the Moon, the jump height is about 6 times higher than on Earth. Hence, the correct option is (2).

Answer:

I think the distance should be given in the question because the formula for getting the

average speed= total distance/total time taken

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Answer:

(i) v = 44 m/s

(ii) a = 72 m/s^2

Explanation:

You have the following equation for the potion of a car:

[tex]x=20t+6t^4[/tex]

(i) The instantaneous velocity is the derivative of x in time:

[tex]\frac{dx}{dt}=20+(6)(4)t^3=20+24t^3[/tex]

for t = 1 is:

[tex]v(t=1)=\frac{dx}{dt}=20+24(1)^3=44m/s[/tex]

(ii) The instantaneous acceleration is the derivative of the velocity:

[tex]\frac{dv}{dt}=24(3)t^2=72t^2[/tex]

for t = 1

[tex]a(t=1)=\frac{dv}{dt}=72(1)^2=72m/s^2[/tex]

Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]     (1)

x: position of the ball after t seconds = 87 m

t: time  = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

[tex]x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}[/tex]

You replace the values of the parameters in the previous equation:

[tex]a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}[/tex]

The acceleration of the ball is 2.35 m/s^2

Answer:

[tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of [tex]25^{\circ}C[/tex]. Then:

[tex]\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}[/tex]

[tex]\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})[/tex]

Finally, the amount of heat released from water is now computed by replacing variables:

[tex]\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right][/tex][tex]\Delta H_{tot} = 2258.025\,kJ[/tex]

An object with momentum must also have impulse.

An object with momentum must also have: c. impulse.

Momentum can be defined as the product of the mass possessed by an object and its velocity. Also, it is a vector quantity and as such has both magnitude and direction.

Mathematically, momentum is giving by the formula;

[tex]Momentum = Mass \times Velocity[/tex]

In Physics, the impulse experienced by an object is always equal to the change in momentum of the object. This is due to the force acting on an object.

[tex]Impulse = Change\;in\;momentum\\\\Force \times time = m \Delta V[/tex]

In conclusion, an object with momentum must also have impulse.

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Answer:

735.75 N

Explanation:

We are given;

Area of small lift; A1 = 5 cm²

Area of large lift; A2 = 300 cm²

Mass required to lift large lift;m = 4500 kg

Since Force = mg, then, Force required to lift large lift;F2 = 4500 × 9.81 = 44145 N

By Pascal's law,

F1/A1 = F2/A2

We want to find minimum force that must be applied to the small piston;F1, so let's make F1 the subject.

Thus;

F1 = (F2 × A1)/A2

Plugging in the relevant values, we have;

F1 = (44145 × 5)/300

F1 = 735.75 N

Answer: horizontal components

Explanation:use the Cardinals point

Answer:

E = 9.62*10^-8 J

Explanation:

The energy stored in a capacitor is given by the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]     (1)

E: energy stored

C: capacitance

V: potential difference of the capacitor = 4 V

The capacitance for a concentric cylindrical capacitor is:

[tex]C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}[/tex]     (2)

L: length of the capacitor = 150m

r2: radius of the outer cylinder  = 8mm = 8*10^-3m

r1: radius of the inner cylinder = 4mm = 4*10^-3m

εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

You replace the expression (2) into the equation (1) and replace the values of all parameters:

[tex]E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J[/tex]

The energy stored in the cylindrical capacitor is 9.62*10-8 J

Before the engines fail, the rocket's altitude at time t is given by

[tex]y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

and its velocity is

[tex]v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

The rocket then reaches an altitude of 1150 m at time t such that

[tex]1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

Solve for t to find this time to be

[tex]t=11.2\,\mathrm s[/tex]

At this time, the rocket attains a velocity of

[tex]v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}[/tex]

When it's in freefall, the rocket's altitude is given by

[tex]y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity, and its velocity is

[tex]v_2(t)=124\dfrac{\rm m}{\rm s}-gt[/tex]

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for [tex]y_2(t)[/tex] to reach 0:

[tex]1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s[/tex]

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]

where [tex]v_f[/tex] and [tex]v_i[/tex] denote final and initial velocities, respecitively, [tex]a[/tex] denotes acceleration, and [tex]\Delta y[/tex] the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means [tex]y_2[/tex] will contain the information we need to find the maximum height.

[tex]-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)[/tex]

Solve for [tex]y_{\rm max}[/tex] and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to [tex]y_2(t)[/tex]) to be about 32.6 s. Plug this into [tex]v_2(t)[/tex] to find the velocity before it crashes:

[tex]v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}[/tex]

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

Answer:

A, D and E

Explanation:

The idea behind elastic is take an elastic band for instance if you stretch it, it will assume it's original shape and form when it returns to its original position hence nothing is lost.

Similarly for elastic collision momentum and kinetic energy are conserved hence; the momentum lost by one is gained by the other and the kinetic energy lost by one is gained by the other.

We have that for the Question ,it can be said that the the appropriate option that identifies elastic collision are

Therefore Option A,D,C

From the question we are told

Which of the following statements is true for an elastic collision?

A. Both momentum and kinetic energy are conserved.

B. Momentum is conserved, but kinetic energy is not conserved.

C. Kinetic energy is conserved, but momentum is not conserved.

D. The amount of momentum lost by one object is the same as the amount gained by the other object.

E. The amount of kinetic energy lost by one object is the same as the amount gained by the other object.

Generally

Elastic Collision

Elastic collision sees an encounter between two bodies have a net kinetic energy between two bodies remains the same.

Therefore

The the appropriate option that identifies elastic collision are

Therefore Option A,D,C

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Answer:

physiology is the study of mechanisms and functions in a living system

Explanation:

it's basically a part of biology that works with how a living organism or body part functions

Answer:

V = 266.23 m/s

Explanation:

The speed of the wave can easily be given by the following formula:

V = S/t

where,

V = Speed of the Wave = ?

S = Distance Covered by Wave = 3920 km

S = Distance Covered by Wave = (3920 km)(1000 m/1 km)

S = Distance Covered by Wave = 3.92 x 10⁶ m

t = Time taken by the wave to cover the distance = 4.09 h

t = Time taken by the wave to cover the distance = (4.09 h)(3600 s/1 h)

t = Time taken by the wave to cover the distance = 14724 s

Therefore,

V = (3.92 x 10⁶ m)/(14724 s)

V = 266.23 m/s

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

Answer:

3.10 m

Explanation:

Given;

Initial speed of ball u = 9.20 m/s

Angle θ = 69°

Horizontal distance from goal d = 4.22m

Resolving the initial velocity into horizontal and vertical components;

The horizontal component of the initial velocity;

uh = ucosθ

Substituting the given values;

uh = 9.2cos69°

uh = 3.30 m/s

The time taken for it to cover the horizontal distance of 4.22 m (to reach the goal);

Time = distance/speed = d/uh

time = 4.22/3.30 =. 1.279s

The time taken to reach the goal is 1.279 seconds.

To determine the height of the ball, we will resolve the vertical component of the initial velocity;

Vertical component of the ball velocity is;

Uv = usinθ

Uv = 9.20sin69°

Uv = 8.59 m/s

Applying the equation of motion;

Height h = ut - 0.5gt^2

Velocity u = Uv = 8.59 m/s

Time t = 1.279s

Acceleration due to gravity g = 9.8 m/s^2

Substituting the values;

h = 8.69(1.279) - 0.5(9.8×1.279^2)

Height h = 3.0988891 = 3.10 m

The height of ball above the release point is 3.10m

Answer:

Explanation:

Given that:

length of side , a = 0.5 m

charge , q = 6.65 mC

length of diagonal , d = 0.5 * sqrt(2)

d = 0.707 m

F is the force due to adjacent particle ,

F1 is the force due to diagonal particle

Now , for the net charge on a particle

Fnet = 2 * F * cos(45) + F1

Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2

Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)

Fnet = 3.05 *10^6 N

the magnitude of net force acting on each particle is 3.05 *10^6 N

part B)

for the direction of particle

d) along the line between the charge and the center of the square outward of the center

The correct answer is C. 0.0896 oz

Explanation:

Both grams and ounces are units use to measure the mass of objects. Additionally, 1 gram represents a smaller amount of mass than the one represented by 1 ounce. Indeed, 1 ounce is equivalent to 28.349 grams and therefore, 1 gram is equivalent to 0.035 ounces ( 1 ounces x 1 gram / 28.349 grams = 0.03527 ounces).

According to this, you can convert grams to ounces by multiplying the amount of grams given by 0.035, which is the equivalent of 1 gram to ounces. This means, to determine how many ounces there are in 2.54 you multiply it by 0.035 which is equivalent to 0.0896 ounces (2.54 x 0.03527 = 0.0896 oz).

Answer:

   W₃ = 3310.49 J ,  W3 = 3310.49 J

Explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

       v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

        v² = 2 a₁ y

        a₁ = v² / 2y

        a₁ = 2.3² / (2 5.90)

        a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

         F -W = m a₁

         F = m a₁ + m g

         F = m (a₁ + g)

         F = 69 (0.448 + 9.8)

         F = 707.1 N

Work is defined by

         W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

          W₁ = 707.1 5.9

           W₁ = 4171.89 J  W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

           F - W = 0

           F = W

           F = m g

           F = 60 9.8

           F = 588 A

the job is

           W² = 588 5.9

            W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

           v² = v₀² + 2 a₂ y

            0 = vo2₀² + 2 a₂ y

            a₂ = -v₀² / 2 y

            a₂ = - 2.3²/2 5.9

            a2 = - 0.448 m / s²

we calculate the force

            F - W = m a₂

            F = m (g + a₂)

            F = 60 (9.8 - 0.448)

            F = 561.1 N

we calculate the work

            W3 = F and

            W3 = 561.1 5.9

          W3 = 3310.49 J

total work

          W_total = W1 + W2 + W3

          W_total = 4171.89 +3469.2 + 3310.49

           w_total = 10951.58 J

Answer:

D

Explanation:

Continuous data is a characteristic of analog signals though been smooth is a subject of question. The smooth signal is after it's been processed. However a wave can still stay continuous defined by what we call duty circle.

Smooth and continuous data is one of the major features of an analog signal.

Analog signal isn't as accurate as the digital signal and it contains minimum

and maximum values which could be positive or negative.

Analog signal comprises of smooth and continuous data. This is explained

by one time-varying quantity representing another time-based variable.In

this type of signal , a variable is an analog of the other.

Analog signals examples include the following : Human voice, Thermometers, analog clocks etc

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Answer:

6.25 m/s

Explanation:

Remember that speed(m/s) = distance(m) / time(s)

Answer:

Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.

Explanation:

Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:

Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.

The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.

At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:

[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],

where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.

Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.

Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.

For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:

[tex]B_1 = B_2[/tex].

[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].

Simplify this equation:

[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].

These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.

Hope it helps

Good luck on your assignment

Answer:

(a) 20 m

(b) 6 m/s²

(c) Between t=0 and t=2, the body moves to the left.

Between t=2 and t=4, the body moves to the right.

Explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

(c) v = 3t² − 6t

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2.  Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.

Answer:

V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:

To find the launch velocity, we use 1st equation of motion.

Vf = Vi + at

where,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)

V₀ = 21.81 m/s

Now, for x-component:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)

V₀ₓ = 10.94 m/s

for y-component:

V₀ₓ = V₀ Sin θ

where,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)

V₀y = 18.87 m/s