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Following an inelastic collision, what can kinetic energy be converted into? Check all that apply.
heat
impulse
momentum
potential energy
sound
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Answer:

(A). The magnitude of the momentum of the marble is 0.004745 kg m/s.

(B). The speed of baseball is 23.0 m/s

Explanation:

Given that,

Mass of marble = 0.0073 kg

Speed = 0.65 m/s

(A). We need to calculate the magnitude of the momentum of the marble

Using formula of momentum

[tex]p = mv[/tex]

Where, m = mass

v = velocity

Put the value into the formula

[tex]p=0.0073\times0.65[/tex]

[tex]p=0.004745\ kg m/s[/tex]

(B). Mass of baseball = 0.136 kg

Momentum of baseball = 3.14 kg m/s

We need to calculate the speed of baseball

Using formula of momentum

[tex]p=mv[/tex]

[tex]v=\dfrac{p}{m}[/tex]

Put the value into the formula

[tex]v=\dfrac{3.14}{0.136}[/tex]

[tex]v=23.0\ m/s[/tex]

Hence, (A). The magnitude of the momentum of the marble is 0.004745 kg m/s.

(B). The speed of baseball is 23.0 m/s

Answer:

n an electrometer, it is built in such a way that its resistance in parallel is extremely high

Ground in a circuit is a reference point from which voltages are measured

all the instruments must be grounded and we must ground ourselves

Explanation:

When you build a voltmeter you have a resistance in parallel with the galvanometer, therefore when measuring the voltage of a circuit, so that there is no effect (load effect) by the voltmeter, a resistance must be much greater than the resistance where it is is measuring.

In an electrometer, it is built in such a way that its resistance in parallel is extremely high in the order of 10¹²Ω, so its load effect is very small and can be measured with high resistance mu

Electric ground in home and industrial installations is a protection system consisting of a metal piece connected to a buried ground electrode.

Ground in a circuit is a reference point from which voltages are measured and is common to all parts of the circuit

In an experiment where an electrometer is used, all the instruments must be grounded and we must ground ourselves, since it must be an instrument where very small voltages are measured at high impedances.

Answer:

8.6602 tons

Explanation:

We first draw the known vector forces.

2fcos30⁰

We have f to be equal to 5tons

Inserting into formula

Σfx = 2(5)cos30⁰

= 8.6602 tons

Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.

Therefore the resultant force on the ship is equal to 8.6602 tons

I hope this helps!

Please check attachment for diagram.

Answer:

learning new activities like riding a bike and playing sports. and also having the ability to collect information and remember them (memory). <3

Explanation:

13+x=43>2x

13+x=43>2x+43>2x

13+x=86>4x

x-4x=86-13

3x=73

x=73/3

x=24.333

x=24.4

Answer:

water displacement.

the bigger the item means more water displacement meaning bouyance

Answer:

10 miles per minute.

Answer:

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Explanation:

We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:

x-Direction

[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]

y-Direction

[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.

[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.

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

If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:

[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]

[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)

[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)

At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):

[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]

[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]

(Ec. 1) in (Ec. 2):

[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]

Now we clear the time in the resulting expression:

[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]

[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]

There are two solutions:

[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

The initial velocity is cleared within (Ec. 2):

[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]

The bounds of the range of initial speed is determined hereafter:

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]

[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]

[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Answer:

The value is [tex]U = 0.06 \ J [/tex]

Explanation:

From the question we are told that

The value of charge on each three point charge is

[tex]q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C[/tex]

The length of the sides of the equilateral triangle is [tex]r = 0.500 \ [/tex]

Generally the total potential energy is mathematically represented as

[tex]U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ][/tex]

=> [tex]U = k * 3 * \frac{q^2}{r} [/tex]

Here k is coulomb constant with value [tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

=>    [tex]U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }  [/tex]

[tex]U  = 0.06 \  J  [/tex]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.

(a) What is the magnitude of the satellite's velocity when the thruster turns off

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ([tex]\vec{v}_{S}[/tex]), measured in meters per second, is:

[tex]\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}[/tex]

Where:

[tex]v_{o,x}[/tex] - Initial velocity in +x direction, measured in meters per second.

[tex]a_{x}[/tex] - Acceleration in +x direction, measured in meter per square second.

[tex]t[/tex] - Time, measured in seconds.

[tex]v_{y}[/tex] - Velocity in +y direction, measured in meters per second.

If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]a_{x} = 0.250\,\frac{m}{s^{2}}[/tex], [tex]t = 45\,s[/tex] and [tex]v_{y} = 21.4\,\frac{m}{s}[/tex], the final velocity of the satellite is:

[tex]\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}[/tex]

[tex]\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right][/tex]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

[tex]\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}[/tex]

[tex]\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}[/tex]

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

[tex]\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }[/tex]

[tex]\tan \alpha = 1.902[/tex]

[tex]\alpha = \tan^{-1}1.902[/tex]

[tex]\alpha \approx 62.266^{\circ}[/tex]

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.

Ionic bonds, also known as electrovalent bonds, are a type of connection created in a chemical molecule by the electrostatic attraction of ions with opposing charges.

As we know,

It has been shown that the top of the electronegativity scale is defined by the fundamental atomic particles F>O>N.

Ionic bonds need an electron, often a nonmetal, and an electron, typically a metal. Metals display ionic bonding because there aren't many-electron in outer orbitals.

Thus, sulfur, a nonmetal with 6 valence electrons atom is most likely to accept electrons to form an ionic bond option (D) correct.

Learn more about the ionic bond here:

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The direction of movement of the body or item is defined by velocity. Your total displacement will be 684 meters.

The direction of movement of the body or item is defined by velocity. Speed is fundamentally a scalar number. Velocity is, in essence, a vector quantity. It is the pace at which distance changes. It is the displacement rate of change.

Given that It takes you 9.5 min to walk with an average velocity of 1.2 m/s to the north from the bus stop to the museum entrance.

Time = 9.5 minutes = 9.5 × 60 seconds = 570 seconds

Average Velocity = 1.2 m/s

Now, the displacement made by you can be written as,

Velocity = Displacement / Time

1.2 m/sec = Displacement / 570 seconds

Displacement = 1.2 m/sec × 570 seconds

Displacement = 684 meter

Hence, your total displacement will be 684 meters.

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Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

P.S. This is part of the answer of another brainly user this is not my knowledge.

But Hope I helped you

Answer:

-0.133 m/s²

Explanation:

Given:

Δx = 541 m

v₀ = 12 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (12 m/s)² + 2a (541 m)

a = -0.133 m/s²

Answer:

Electric current in amperes = 1.1808 A

Explanation:

Given:

Intercept protons rate = 1800 protons per second

Area = 41 × 10⁷ km²

Find:

Electric current in amperes

Computation:

Current density = Intercept protons rate × 1.6 × 10⁻¹⁹

Current density = 1800 × 1.6 × 10⁻¹⁹

Current density = 2.88 × 10⁻¹⁶

1 km² = 10⁶m²

So,

Electric current in amperes =  2.88 × 10⁻¹⁶ × 41 × 10⁷ × 10⁶

Electric current in amperes = 1.1808 A

Answer:

1. Golf ball is in the range of only particulate detectable range

2, alpha partcle is in ultraviolet range of wavelength

3.bullet is in xray range of wavelength

All in the EMW spectrum

Answer:

[tex]A=90^0-B\\B+C=180^0-90^0=90^0\\C=90^0-B\\A=C[/tex]

Explanation:

Answer:

I know

Explanation:

Physics= hard

Answer:

Please add the multiple choice awensers too then I'll get back to you when I can

Answer:

Answer is D

Explanation:

Answer:

The area is [tex]A = 0.015 \ m^2 [/tex]

The resistance is  [tex]R =  1.0 *10^{5} \ \Omega [/tex]

Explanation:

From the question we are told that

The relativity of dry skin is [tex]\rho = 1 * 10^6\ \Omega\cdot m[/tex]

The thickness of the skin is [tex]d = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

Generally the area of a flat palm with length = 0.10 m and breath b = 0.15 m is mathematically represented as

[tex]A = d * L[/tex]

[tex]A = 0.10 * 0.15 [/tex]

[tex]A = 0.015 \ m^2 [/tex]

Generally the resistance is mathematically represented as

[tex]R = \frac{\rho * d}{ A }[/tex]

=>    [tex]R =  \frac{ 1.0 *10^{6} *   1.5*10^{-3}}{ 0.015  }[/tex]

=> [tex]R = 1.0 *10^{5} \ \Omega [/tex]

Answer:

80

Explanation:

because of the denstiy formula

Answer:

224.6 N

Explanation:

We can use first the formula to calculate the centripetal acceleration, given by:

[tex]a_c=\frac{v_t^2}{R}[/tex]

where the Vt is the tangential velocity, and R is the radius of the circular motion.

Then, for our case we have:

[tex]a_c=\frac{v_t^2}{R}=\frac{3.5^2}{0.6}\approx 20.42\,\,\frac{m}{s^2}[/tex]

And now we multiply this acceleration by Miguel's mass (11 kg) to obtain the centripetal force acting on him:

[tex]F_c=11 \,*\,20.42\,N = 224.6\,\,N[/tex]

Answer:

225 N

Explanation:

Just answered the question

The paper plane slows down when flying in the air due to air resistance or the friction due to the air around it. As the paper plane have flat surfaces and more surface area, the air resists it's motion more than before. If the paper plane is moving towards left, then friction acts towards left.

The friction always act opposite to the direction of paper plane and resists its motion. Hence, we can say that paper plane slows due to Frictional Force [tex]\red{\bullet}[/tex]

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Answer:

F=200N

a=500m/s2​

Mass=?

Explanation:

F=ma

200=m*500

200/500=m

Mass=0.4kg

Answer:

[tex]x=8m[/tex]

Explanation:

Hello,

In this case, the most suitable kinematic equation for the calculation of the squirrel's displacement is:

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

Whereas the initial velocity is zero as it starts from rest, thus, we obtain:

[tex]x=\frac{1}{2}*1.0\frac{m}{s^2}*(2.0s)^2\\ \\x=8m[/tex]

Best regards.

Responder:

Explicación:

Usaremos la ecuación de movimiento para determinar la altura de la bola medida desde la parte superior del edificio.

Usando la ecuación para obtener la altura de caída

S = ut + 1 / 2gt²

u es la velocidad inicial = 25 m / s

g es la aceleración debida a la gravedad = 9,81 m / s²

t es el tiempo = 7 segundos

S es la altura de la caída

S = 25 (7) +1/2 (9,81) × 7²

S = 175 + 4,905 (49)

S = 175 + 240,345

S = 415,35 m

Esto significa que la pelota se elevó a 415,35 m de altura

La altura vertical máxima alcanzada es de 31,88 m.

Tenemos la siguiente información de la pregunta;

Velocidad inicial = 25 m/s

Velocidad final = 0 m/s (a la altura máxima)

tiempo empleado = 3,5 minutos (el tiempo empleado para subir y bajar es igual).

Usando la ecuación;

v^2 = u^2 - 2gh

Dado que v = 0

u^2 = 2gh

h = tu^2/2g

h = (25)^2/2 *9.8

h = 31,88 m

Obtenga más información sobre las ecuaciones de movimiento: https://brainly.com/question/8898885

Answer:

A

Explanation:

Answer:

2 seconds

Explanation:

The time taken for the movement is 2.81 seconds.

To find the time, the given values are,

Distance = 45 meters

speed = 16 m/s.

Formula of distance,

              Distance = Speed × Time.

As we have to find the time, Formula can be rearranged as,

              Time = Distance / Speed

Substituting the given datas in the formula,

Time = 45 / 16

        = 2.81 seconds.

So, the time taken for the movement is 2.81 seconds.

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