A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.

Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

Voltage of the battery, V = 25.5 V

We need to find the power supplied to the starter motor. We know that,

P = VI

Put all the values,

P = 25.5 × 260

P = 6630 W

or

P = 6.63 kW

So, the power supplied to the motor is 6.63 kW.

Answer:

The power is 6.63 kW.

Explanation:

Current, I = 260  A

Voltage, V = 25.5 V

Power of an electrical appliance is given by

P = V I

P = 25.5 x 260

P = 6630 W

1 kW = 1000 W

So, the power is

P = 6.63 kW

Answer:

XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time

Explanation:

so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be

Answer:

b.have frequencies that are integer multiples of the frequency of the complex waveform

Explanation:

Please correct me if I am wrong

Answer:

100j

Explanation:

Explanation:

Displacement of a body moving in circular motion is called uniform circular motion.

hope it is helpful to you

Answer:

A constantly moving object with consistent circular movement. However, for its change in direction, it is accelerating

Explanation:

Uniform circular motion in a circle at constant rate can be described as the motion of the object. When an object moves in a circle, it changes its direction constantly. The object moves tangently to the circle at all times. As the velocity vector direction is the same as the object motion direction, the velocity vector is tangent to the circle. This is shown in the animation on the right by a vector arrow.

An item is accelerating that moves in a circle. Objects that accelerate are subjects that change their speed – either the velocity (i.e. the vecteur magnitude) or the direction. An object with consistent circular movement moves at a constant speed. However, because of its change in direction, it is accelerating. The acceleration direction is inside. The animation on the right shows this through a vector arrow

For an object with only a uniform circular movement, the final motion is the net force. The The net force acting on this object is directed to the middle of the circle. The net force is an inner or centripetal force. Without such a deepest force, an object would continue in a straight direction, never deviating. Regrettably, with the inward net force, perpendicular to the vector, the object changes the direction and is accelerated internally.

Answer: The focal length of the lens is 2.60 cm

Explanation:

The equation for lens formula follows:

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

where,

f = focal length = ? cm

v = image distance = 2.9 cm

u = Object distance = -25 cm

Putting values in above equation, we get:

[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]

Hence, the focal length of the lens is 2.60 cm

Answer:

Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.

Explanation:

Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.

Answer:

The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).

Explanation:

We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.

P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.

Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node

Answer:

f = 7.06 N

Explanation:

The maximum frictional force on the knee joint of the person can be given by the following formula:

[tex]f = \mu R = \mu W \\[/tex]

where,

f = maximum frictional force = ?

μ = static friction coefficient = 0.016

W = Weight load on knee = mg

m = mass supported by knee = 45 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]

f = 7.06 N

Answer:

1140 J, 6840 J, 10260 J

Explanation:

Lo x 2 Lo x 3 Lo, Lo = 0.2 m,  K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,

time, t = 6 s

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]

Answer:

A) v_{f1} = -3.2 m / s,  B) LEFT , C) v_{f2} = -0.12 m / s,  D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

          p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

          p_f = m₁ v_{f1} + m₂ v_{f2}

          p₀ = p_f

          m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

 as the shock is elastic, energy is conserved

         K₀ = K_f

         ½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]

         m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

         (a + b) (a-b) = a² -b²

         m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

         m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂)                                  (1)

         m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

         v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

         M = m₁ + m₂

         [tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]

         [tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

         m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

         v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]

         v_{f1} = -0,250 - 2,961

          v_{f1} = - 3,211 m / s

         v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]

         v_{f2} = 0.570 - 0.6909

         v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

Answer:

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]

From the image, expressing the forces through the y-axis, we have:

[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]

Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:

Then:

[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]

[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]

[tex]F_2 = 117600 \ N[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

For the force equation:

[tex]F_1+F_2=1.617*10^5 \ N;[/tex]

where:

[tex]F_2 = 1.176*10^5 \ N[/tex]

Then:

[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]

[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]

[tex]F_1=44100\ N[/tex]

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

Answer:

 T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Explanation:

Let's use Newton's second law

          F = ma

force is the universal force of attraction and acceleration is centripetal

          G m M / r² = m v² / r

          G M / r = v²

as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion

          v = d / T

the length of a circle is

          d = 2π r

we substitute

        G M / r = 4π² r² / T²

        T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]

the distance r is measured from the center of the Earth (Re), therefore

        r = Re + h

where h is the height from the planet's surface

let's calculate

         T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex]   (Re + h) ³

         T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]

         T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Explanation:

The initial kinetic energy [tex]KE_0[/tex] is

[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]

When its mass and velocity are doubled, its new kinetic energy KE is

[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]

[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]

Therefore the kinetic energy will increase by a factor of 8.

Answer:

6.2 × 10^-11 m

Explanation:

1 eV = 1.602 × 10-19 joule

2.00 × 104 ev. = 2.00 × 10^4 eV × 1.602 × 10^-19 joule/1eV

= 3.2 × 10^-15 J

From;

E= hc/λ

λ = hc/E

λ = 6.6 × 10^-34 × 3 × 10^8/3.2 × 10^-15

λ = 6.2 × 10^-11 m

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Answer:

i think the answer is 0.001m³

Answer:

2.70

Explanation:

pH = -log[H+]

pH = -log[2.0x10^-3]

pH = 2.70

Full Question:

What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?

A) 0°F

B) 273 K

C) 0 K

D) 100°C

E) 273°C

Answer:

The correction Option is D) 100°C

Explanation:

The temperature above is referred to as the critical point.

it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.

There is also a condition under which water can exist in its three forms: that is  

- Ice (solid)

- Liquid (fluid)

- Gas (vapor)

That state is called triple point. The conditions necessary for that to occur are:

Cheers

Cheers

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

The question is incomplete. The complete question is :

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].

Determine the following:

(a) frequency of the motion

(b) period of the motion

(c) amplitude of the motion

(d) first time after t = 0  that the object reaches the position x = 2.6 cm

Solution :

Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].

Comparing it with the general equation of simple harmonic motion,

 x = A sin (ωt + Φ)

  A = 4.7 cm

  ω = 7.9 π

a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]

                                             [tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]

                                             = 3.95 Hz

b). The period, [tex]$T=\frac{1}{f}$[/tex]

                        [tex]$T=\frac{1}{3.95}[/tex]

                            = 0.253 seconds

c). Amplitude is A = 4.7 cm

d). We have,

    x = A sin (ωt + Φ)

    [tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]

    [tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]

     [tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]

     [tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]

          Hence, t = 0.0236 seconds.

Answer:

-45,597.07

Explanation:

if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh

Answer:

Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1

Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2

(Xcm , Ycm) = (1 , 1/2)

Using definition of center of mass

Answer:

p = 1.2 kg-m/s

Explanation:

The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".

Mass of the object, m = 0.5 kg

Acceleration of the object, a = 0.3 m/s²

We need to find the momentum after 8 seconds.

We know that,

[tex]p=F\times t[/tex]

i.e.

p = mat

So,

[tex]p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s[/tex]

So, the momentum of the object is 1.2 kg-m/s.

Answer: [tex]29.85\ W/m^2[/tex]

Explanation:

Given

Power [tex]P=60\ W[/tex]

Distance from the light source [tex]r=0.4\ m[/tex]

Intensity is given by

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Inserting values

[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]

Answer:

29.85 W/ m^2

Explanation:

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.

1 eV = 1.60 * [tex]10^{-19} J[/tex]     work done in accelerating electron throw 1 V

K.E (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules

1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J

Velocity of proton is,

v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]

v = 5.09 * [tex]10^{5}[/tex]m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

To learn more about velocity refer the link:

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Answer:

C.  Rotational motion

Explanation:

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).

Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.

"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."

Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.

Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.

The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.

Learn more about motion here:

https://brainly.com/question/22810476

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Answer:

wareffctgggyyggghhhh

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.