A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.40 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum (Let up be in the positive direction.)

Answer:

The car travels 100 meter during its 5.0 s of acceleration

Explanation:

We have the equation of motion v²=u²+2as, where v is the final velocity, u is the initial velocity, a is the acceleration ans s is the displacement.

v = 25 m/s

a = 2 m/s²

u = 15 m/s

Substituting

      25²=15²+2 x 2 x s

       s = 100 m

The car travels 100 meter during its 5.0 s of acceleration

Answer:

[tex]0.6 \times 10 {}^{15} [/tex]

Explanation:

By using relation,

Speed of light = frequency × wavelength (in m)

Answer:

2.62 m

Explanation:

Let the small child sit at a distance x from the pivot.

The distance of big child from the pivot is 4 - x .

By using the concept of moments.

Clockwise moments = anticlockwise moments

27 x = 51 ( 4 - x )

27 x = 204 - 51 x

78 x = 204

x = 2.62 m

Answer:

catch the ball

Explanation:

if you push the ball back, that equal amount of force will applied to you. however if you catch it, you absorb less of the directional energy

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

Answer:

Explanation:

When a body is moving in a liquid, it experiences a backward dragging force. this backward dragging force is given by the formula

F = 6 π η r v

where, η is called the coefficient of viscosity, r be the radius, v be the velocity of object.

As we know that the coefficient of friction for oil is more than water so the dragging force in oil is more than water.

So, the balls in water comes first.

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

[tex]v^2=u^2+2as[/tex]

Here final velocity, v = 26 m/s

a = acceleration due to gravity

[tex]a=9.8m/s^2 \\ [/tex]

displacement, s = 33 m

Substituting

[tex]26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\ [/tex]

Speed of water at the top of fall = 5.40 m/s

Answer:

Resistance of the bulb is 2100 watts.

Explanation:

Given that,

Power of the bulb, P = 7.5 watts

Voltage, V = 125 volts

We have to find the resistance of the bulb. The power of an electrical appliance is given by the following formula as :

[tex]P=\dfrac{V^2}{R}[/tex]

[tex]R=\dfrac{V^2}{P}[/tex]

[tex]R=\dfrac{(125\ V)^2}{7.5\ W}[/tex]

R = 2083.34 ohms

or

R = 2100 ohms

Hence, the correct option is (d) "2100 ohms"

Answer:

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

Explanation:

Initial momentum of the ball is given as

[tex]P_i = mv_i[/tex]

[tex]P_i = 0.145 (40) = 5.8 kg m/s \hat i[/tex]

now final momentum of the ball is given as

[tex]P_f = 0.145(57) = 8.3 kg m/s \hat j[/tex]

now by the formula of force we have

[tex]F = \frac{P_f - P_i}{\Delta t}[/tex]

now we have

[tex]F = \frac{8.3 \hat j - 5.8 \hat i}{2.10 \times 10^{-3}}[/tex]

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

The average vector force the ball exerts on the bat during their interaction is

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

Given :

Initial Speed = 40 m/sec

Final Speed = 57 m/sec

Mass = 0.145 Kg

Time = 0.0021 sec

Solution :

Initial Momentum is,

[tex]\rm P_i = mv_i[/tex]

[tex]\rm P_i = 0.145\times 40 = 5.8\;Kg .m/sec\;\hat{i}[/tex]

Final momentum is,

[tex]\rm P_f = mv_f = 0.145\times57=8.3\; Kg.m/sec\; \hat{j}[/tex]

Now,

[tex]\rm F=\dfrac{P_f-P_i}{\Delta t}= \dfrac{8.3\hat{j}-5.8\hat{i}}{2.10\times10^-^3}[/tex]

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

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Answer:

48 m/s

Explanation:

m = mass of the object = 28 kg

F = magnitude of net force acting on the object = 232 N

acceleration of the object is given as

a = F/m

a = 232/28

a = 8.3 m/s²

v₀ = initial velocity of the object = 5 m/s

v = final velocity of the object

t = time interval = 5.2 s

using the kinematics equation

v = v₀ + a t

v = 5 + (8.3) (5.2)

v = 48 m/s

The power output of a 1.5 g fly moving upwards at a speed of 2.4 cm/s can be calculated by applying the physics formula for power. The resultant power output is ~0.036 Watts.

The power output of any moving body can be calculated using the formula P = mgh/t where 'P' is power, 'm' is mass, 'g' is acceleration due to gravity, 'h' is height and 't' is time. In order to calculate the power output of the fly, we need to convert the variables to the appropriate units.

So first, convert the mass of the fly to kg, which gives 0.0015 kg. Then, convert the speed from cm/s to m/s, giving us 0.024 m/s. The acceleration due to gravity is approximately 9.8 m/s2. We want to find the power as the fly walks 1 meter.

Substituting the values into the power equation, we get P = (0.0015 kg * 9.8 m/s2 * 1 m) / (1 m / 0.024 m/s) = 0.036 W, which when rounded off to two significant figures is 0.036 Watts.

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To calculate the power output of a fly walking up a windowpane, we need to use the equation Power = force x velocity. By converting the mass of the fly into kilograms, calculating the force exerted by the fly using the equation Force = mass x gravity, and multiplying the force by the velocity, we find that the power output of the fly is approximately 0.00004 Watts or 4 x 10^-5 Watts.

To calculate the power output of the fly, we can use the equation: Power = force x velocity.

First, let's convert the mass of the fly from grams to kilograms. 1.5 g = 0.0015 kg. The force exerted by the fly is equal to its weight, which is given by the equation: Force = mass x gravity.

Assuming the acceleration due to gravity is 9.8 m/s^2, the force exerted by the fly is: Force = 0.0015 kg x 9.8 m/s^2. Finally, we can calculate the power output of the fly by multiplying the force by the velocity: Power = (0.0015 kg x 9.8 m/s^2) x 0.024 m/s. This gives us a power output of approximately 0.00004 Watts or 4 x 10^-5 Watts.

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Answer:

0.186 N/C

Explanation:

The relationship between electric field strength and potential difference is:

[tex]E=\frac{\Delta V}{d}[/tex]

where

E is the electric field strength

[tex]\Delta V[/tex] is the potential difference

d is the distance

Here we have

[tex]\Delta V=41 mV=0.041 V[/tex]

d = 22 cm = 0.22 m

So the electric field magnitude is

[tex]E=\frac{0.041 V}{0.22 m}=0.186 N/C[/tex]

Final answer:

To find the electric field inside a conductor, use the formula E = V/d. With a potential difference of 41 mV and length of 22 cm, the electric field is 0.186 N/C.

Explanation:

The magnitude of the electric field inside a conductor is calculated using the formula E = V/d, where V is the potential difference and d is the length across which the potential difference is applied.

In this case, the conducing rod has a length of 22 cm (0.22 meters) and a potential difference across its ends of 41 mV (0.041 volts). Therefore, the magnitude of the electric field E in the conductor can be calculated as:

E = V/d = 0.041 V / 0.22 m = 0.18636... V/m

Since 1 V/m is equivalent to 1 N/C, the magnitude of the electric field in the conductor is approximately 0.186 N/C.

Answer: [tex]4.068(10)^{13} km[/tex]

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.

This unit is equivalent to [tex]9.461(10)^{12}km[/tex], which mathematically is expressed as:

[tex]1Ly=9.461(10)^{12}km[/tex]

Doing the conversion:

[tex]4.3Ly.\frac{9.461(10)^{12}km}{1Ly}=4.068(10)^{13}km[/tex]

The Answer: The star closest to our sun is 4.2 light years away from the sun.

Explanation:

The frequency of the wave is 4921.57 Hz. The angular frequency is 30937.48 rad/s. The period of oscillation is 0.000203 s.

To find the frequency of the wave, we can use the formula:

v = λ × f

Plugging in the values given, we have:

251 = 0.051 × f

solving for f, we get:

f = 251 / 0.051 = 4921.57 Hz

The angular frequency (represented as ω) can be calculated using the formula:

ω = 2π × f

Substituting the value of f we found, we get:

ω = 2×3.1416×4921.57 = 30937.48 rad/s

The period of oscillation (represented as T) is the reciprocal of the frequency.

T = 1 / f = 1 / 4921.57 = 0.000203 s

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Answer:

Pendulum B

Explanation:

The time period of a pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Case 1.

Mass, m = 200 g = 0.2 kg

Length of string, l = 1 m

Time, [tex]T_1=2\pi\sqrt{\dfrac{1\ m}{9.8\ m/s^2}}[/tex]

T₁ = 2.007 Seconds

Since, [tex]f=\dfrac{1}{T_1}[/tex]

[tex]f_1=\dfrac{1}{2.007}[/tex]

f₁ = 0.49 Hz

Case 2.

Mass, m = 400 g = 0.4 kg

Length of string, l = 0.5 m

Time, [tex]T_2=2\pi\sqrt{\dfrac{0.5\ m}{9.8\ m/s^2}}[/tex]

T₂ = 1.41 seconds

[tex]f₂=\dfrac{1}{T_2}[/tex]

[tex]f₂=\dfrac{1}{1.41}[/tex]

f₂ = 0.709 seconds

Hence, pendulum B have highest frequency of vibration.

Answer:

Radius = 0.11 m

Explanation:

To find the speed of the proton we know that

[tex]KE = PE[/tex]

here we have

[tex]\frac{1}{2}mv^2 = qV[/tex]

now we have

[tex]v = \sqrt{\frac{2qV}{m}}[/tex]

now we have

[tex]v = \sqrt{\frac{2(1.60 \times 10^{-19})(1000)}{(1.67\times 10^{-27})}}[/tex]

[tex]v = 4.38 \times 10^5 m/s[/tex]

Now for the radius of the circular motion of charge we know

[tex]\frac{mv^2}{R} = qvB[/tex]

[tex]R = \frac{mv}{qB}[/tex]

[tex]R = \frac{(1.67\times 10^{-27})(4.38 \times 10^5)}{(1.60\times 10^{-19})(0.040)}[/tex]

[tex]R = 0.11 m[/tex]

The radius of the proton orbit is 0.114m

When the proton travels through the potential V it gains kinetic energy given below:

[tex]\frac{1}{2}mv^2=qV\\\\v=\sqrt[]{\frac{2qV}{m} }\\\\v=\sqrt{\frac{2\times (1.6\times10^{-19})\times10^3}{1.67\times10^{-27}}[/tex]

[tex]v=4.37\times10^5m/s[/tex]

Now, a moving charge under magnetic field B undergoes circular motion due to the magnetic force being perpendicular to the velocity of the charge, given by:

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

[tex]r=\frac{mv}{qB} \\\\r=\frac{1.67\times10^{-27}\times4.37\times10^5}{1.6\times10^{-19}\times0.040} m\\\\r=0.114m[/tex]

the radius of the orbit is 0.114m

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Answer:

The radius of a circular path of a charged particle orbit depends on the charge and velocity of the particle.

Explanation:

Answer:

The radius of the disc is 2.098 m.

(e) is correct option.

Explanation:

Given that,

Moment of inertia I = 12100 kg-m²

Mass of disc m = 5500 kg

Moment of inertia :

The moment of inertia is equal to the product of the mass and square of the radius.

The moment of inertia of the disc is given by

[tex]I=\dfrac{mr^2}{2}[/tex]

Where, m = mass of disc

r = radius of the disc

Put the value into the formula

[tex]12100=\dfrac{5500\times r^2}{2}[/tex]

[tex]r=\sqrt{\dfrac{12100\times2}{5500}}[/tex]

[tex]r= 2.098\ m[/tex]

Hence, The radius of the disc is 2.098 m.

The electric force is a force that is inversely proportional to the square of the distance. This can be proved by Coulomb's Law, which states:  

"The electrostatic force [tex]F_{E}[/tex] between two point charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is proportional to the product of the charges and inversely proportional to the square of the distance [tex]d[/tex] that separates them, and has the direction of the line that joins them"  

Mathematically this law is written as:  

[tex]F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}[/tex]

Where [tex]K[/tex] is a proportionality constant.

So, if we have two electrons, this means we have two charges with the same sign, hence the electric force between them will be repulsive.

Final answer:

Explaining the attractive electrostatic force and gravitational force between two electrons at atomic scales.

Explanation:

The electrostatic force between two electrons separated by 10^-10 m is attractive. The magnitude of this force can be calculated using Coulomb's law, and for two electrons with charge e = 1.6 x 10^-19 C, the force is strong at atomic distances.

The gravitational force between the electrons is always attractive and weaker than the electrostatic force, with a magnitude of 5.54 x 10^-51 N. This force is much smaller compared to the electrostatic force at atomic scales.

The balance between attractive and repulsive forces is crucial in understanding the stability of atomic structures and the concept of bond length and bond energy in molecules.

Answer:

0.36 kg

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum at the beginning must be equal to the total momentum after the skater has thrown the ball.

Before the launch, the skater and the snowball are at rest, so the initial total momentum is zero:

[tex]p_i = 0[/tex]

After the launch, the total momentum is:

[tex]p_f = M V + m v[/tex]

where

M = 53.4 kg is the mass of the ice skater

v = -0.100 m/s is the velocity of the ice skater (here we assumed that east is the positive direction)

m is the mass of the snowball

v = +14.8 m/s is the velocity of the snowball

Since momentum must be conserved,

[tex]p_i = p_f\\0 = MV +mv[/tex]

so we can find m:

[tex]m=-\frac{MV}{v}=-\frac{(53.4 kg)(-0.100 m/s)}{14.8 m/s}=0.36 kg[/tex]

The question pertains to the principle of conservation of momentum. The mass of the snowball is found by equating the momentum of an ice skater with the snowball, resulting in a mass of about 0.36 kg for the snowball.

This question represents a physics concept known as conservation of momentum, stating that the total initial momentum of a system equals the total final momentum in the absence of external forces. In this scenario, the initial momentum of the system is zero because both the ice skater and the snowball are initially at rest.

In the final state, the 53.4 kg ice skater moves west at 0.100 m/s, and the snowball moves east at 14.8 m/s. The eastward momentum equates to the westward momentum.

To find the mass of the snowball, we can set the two momenta to be equal, resulting in the equation: (Mass of Skater) * (Velocity of Skater) = (Mass of Snowball) * (Velocity of Snowball), which simplifies to (53.4 kg) * (0.100 m/s) = (Mass of Snowball) * (14.8 m/s). Solving for the mass of the snowball gives approximately 0.36 kg.

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Answer:

26.7 m/s

Explanation:

The speed of the car is

v = 60 mi/h

We know that

1 mile = 1.6 km = 1600 m

1 h = 60 min = 3600 s

So we can convert the speed from mi/h into m/s by multiplying by the following factor:

[tex]v = 60 \frac{mi}{h} \cdot \frac{1600 m/mi}{3600 s/h}=26.7 m/s[/tex]

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Where:

[tex]V_{o}=200m/s[/tex] is the bullet's initial speed

[tex]\theta=0[/tex] because we are told the bullet is shot horizontally

[tex]t[/tex] is the time since the bullet is shot until it hits the ground

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (2)

Where:

[tex]y_{o}=1.5m[/tex]  is the initial height of the bullet

[tex]y=0[/tex]  is the final height of the bullet (when it finally hits the ground)

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

[tex]0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}[/tex]   (3)

[tex]0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}[/tex]   (4)

Finding [tex]t[/tex]:

[tex]t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}[/tex]   (5)

Then we have the time elapsed before the bullet hits the ground:

[tex]t=0.553s[/tex]   (6)

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Substituting the knonw values and the value of [tex]t[/tex] found in (6):

[tex]x=200m/s.cos(0)(0.553s)[/tex]   (7)

[tex]x=200m/s(0.553s)[/tex]   (8)

Finally:

[tex]x=110.656m[/tex]  

The bullet will hit the ground after approximately 0.553 seconds, and in that time, it will travel horizontally around 110.6 meters.

To determine how much time elapses before the bullet hits the ground (1.5 m drop) when fired horizontally at 200 m/s, we use the equation for free fall motion h = (1/2)gt², where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time in seconds. Solving for time t, we find that the bullet will hit the ground after approximately 0.553 seconds.

For part (b), to find how far the bullet travels horizontally, we simply multiply the time in the air by the bullet's initial horizontal velocity. This gives a horizontal distance of 200 m/s * 0.553 s = 110.6 meters. Therefore, the bullet travels roughly 110.6 meters before hitting the ground.

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Answer:

The ball is at its maximum height.

Explanation:

As we know that when ball is thrown at some angle with the horizontal then the component of its velocity is given as

[tex]v_x = vcos\theta[/tex]

[tex]v_y = vsin\theta[/tex]

now here vertical velocity is the velocity in y direction

so it is given as

[tex]v_y = 3.0 sin30 = 1.5 m/s[/tex]

now as the velocity of ball in vertical direction becomes zero

then in that case

[tex]v_f - v_i = at[/tex]

[tex]0 - 1.5 = (-9.8)t[/tex]

[tex]t = 0.15 s[/tex]

since at this position the vertical component of the velocity is zero

so this is the position of ball when its height is maximum

Final answer:

The vertical velocity of a ball thrown at an angle becomes zero at its maximum height. This is the point in its trajectory where gravity has completely counteracted its initial upward velocity. At no other time during its flight will the ball's vertical velocity be zero.

Explanation:

The motion of a projectile launched at an angle involves two components of motion – horizontal and vertical. When a ball is thrown with a velocity of 3.0 m/s at an angle of 30° above horizontal, it will have both horizontal and vertical components of velocity. The vertical velocity of the ball becomes zero when the ball reaches its maximum height, as there is no upward velocity to counteract gravity at this point.

To find the vertical component of initial velocity (Vy), we use the formula Vy = V × sin(θ), where V is the initial velocity and θ is the launch angle. So in this case, the vertical velocity component is Vy = 3.0 m/s × sin(30°).

At maximum height, gravity has slowed the vertical velocity to zero. This is the only point during the ball's trajectory where the vertical component of velocity is zero. Just before the ball hits the ground, its vertical velocity is not zero but is equal in magnitude and opposite in direction to its vertical velocity on launch. When the ball changes direction horizontally, this does not affect the vertical velocity component. Therefore, the ball's vertical velocity is zero only when it reaches its highest point.

Answer:

When the angle is 56° the work done is 117.43 J

Explanation:

Work = F . s = Fscosθ

We have

   W1 = 210 J, θ = 0°

Substituting

      210 = F x s x cos 0 = Fs

Now we have to find W2 when angle θ = 56°

Substituting

      W2 = F x s x cos 56 = 210 cos56= 117.43 J

When the angle is 56° the work done is 117.43 J

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Answer:

9680 cal

Explanation:

A = cross-sectional area of the bar = 30 cm² = 30 x 10⁻⁴ m²

L = length of the bar = 1.5 m

T₁ = Temperature at one end of the bar = 25 °C

T₂ = Temperature at other end of the bar = 300 °C

k = Thermal conductivity of Aluminum = 1.76 x 10⁴ Cal cm /(m² ⁰C)

Q = amount of heat conducted per second

Amount of heat conducted per second is given as

[tex]Q = \frac{k A (T_{2} - T_{1})}{L}[/tex]

Q = (1.76 x 10⁴) (30 x 10⁻⁴) (300 - 25)/1.5

Q = 9680 cal

The particle has constant acceleration according to

[tex]\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k[/tex]

Its velocity at time [tex]t[/tex] is

[tex]\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du[/tex]

[tex]\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t[/tex]

[tex]\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k[/tex]

Then the particle has position at time [tex]t[/tex] according to

[tex]\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du[/tex]

[tex]\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k[/tex]

At at the point (3, 6, 9), i.e. when [tex]t=0[/tex], it has speed 8, so that

[tex]\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64[/tex]

We know that at some time [tex]t=T[/tex], the particle is at the point (5, 2, 7), which tells us

[tex]\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}[/tex]

and in particular we see that

[tex]v_{0y}=-2v_{0x}[/tex]

and

[tex]v_{0z}=-v_{0x}[/tex]

Then

[tex]{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3[/tex]

[tex]\implies v_{0y}=\mp\dfrac{8\sqrt6}3[/tex]

[tex]\implies v_{0z}=\mp\dfrac{4\sqrt6}3[/tex]

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

[tex]\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k[/tex]

To find the equation for the position vector of the particle, use the formula r(t) = r(0) + v(0) * t + (1/2) * a * t^2.

To find the equation for the position vector of the particle, use the following steps:

For the given problem, the equation for the position vector of the particle is: r(t) = (3 + 4t)i + (6 - 4t + 2t^2)j + (9 - 2t + t^2)k.

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force  

[tex]F = \mu mg[/tex]

[tex]F=0.21\times2.0\times9.8[/tex]

[tex]F = 4.12\ N[/tex]

We need to calculate the acceleration

[tex]F = ma[/tex]

[tex]a = \dfrac{F}{m}[/tex]

[tex]a =\dfrac{4.12}{2.0}[/tex]

[tex]a=2.06\ m/s^2[/tex]

We need to calculate the initial velocity

Using equation of motion

[tex]v^2=u^2-2as[/tex]

Put the value  

[tex]0=15^2-2\times2.06\times s[/tex]

[tex]s = \dfrac{15^2}{2\times2.06}[/tex]

[tex]s=54.6\ m[/tex]

Hence, The distance will be 54.6 m.

Explanation:

Centripetal force is mass times centripetal acceleration:

F = m v² / r

If force is doubled while mass and radius are held constant, then velocity will increase.

2F = m u² / r

2 m v² / r = m u² / r

2 v² = u²

u = v√2

So the velocity increases by a factor of √2.

Doubling the centripetal force acting on an object in circular motion, while maintaining the same radius, necessitates an increase in the object's velocity, leading to faster circular motion.

The question explores how doubling the centripetal force acting on an object in uniform circular motion, while maintaining the same radius, affects its motion. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for centripetal force is given as Fc = m(v2/r), where m is mass, v is velocity, and r is the radius of the circular path. Doubling the centripetal force while keeping the radius constant means that for the force to remain balanced and the object to stay in circular motion, the velocity of the object must increase. This is because the square of the velocity (v2) is directly proportional to the force applied. Therefore, the object's speed around the circular path will increase, resulting in a faster circular motion.

Answer:

(bruh moment)

Explanation: