A car drives on a highway with a speed of 68mi/hr. What is the speed in km/hr?

Answer:

Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.

Explanation:

As we know that the potentiometer can act as a variable resistor but resistors have constant value.

In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.

Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.

Answer:D-velocity

Explanation:

Given

Constant linear acceleration i.e. acceleration is constant

a=constant

and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]

Thus velocity is changing uniformly to give constant acceleration

While displacement and distance are function of time thus they are not changing uniformly.

During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.

Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.

Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.

Therefore,

During constant linear acceleration, the following quantity changes uniformly: velocity.

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Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

[tex]1\ cm=\frac{1}{100}\ m[/tex]

[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]

967500000 years will pass before the radius of the orbit increases by 10%.

Final answer:

The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.

Explanation:

The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.

Hi!

Letw call A to the initially neutral rod, and B to the positively charged. When they are close to each other, the positive charges in B attract the negative charges in A, and repelle the positive ones. If you ground A, negative charges from ground (your body, in this case), flow to A attracted by the positive charges in B, and positive charges in A flow to ground, so finally A results negatively charged

Answer:

Negative charges

Explanation:

The procedure described above is known in physics as charging by electrostatic induction. If we desire to impart negative charges to a hitherto neutral rod, we bring a positively charged rod near it without allowing the two insulated rods to touch each other. If the neutral rod is earthed, negative charges remain on the rod.

Answer:

Explanation:

So, lets say the runner stars from the position [tex]x_0[/tex]. Lets make this point the origin of a coordinate system in which the vector i points north.

[tex]x_0 = (0,0)[/tex]

Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:

[tex]x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)[/tex].

[tex]x_1 = (20 \ m,0)[/tex].

The, he runs 30 meters south

[tex]x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)[/tex]

[tex]x_2 = (-10 \ m,0)[/tex]

Finally, he runs 40 meter north

[tex]x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)[/tex]

[tex]x_3 = (30 \ m,0)[/tex].

This is our displacement vector. Now, the average speed will be:

[tex]\frac{distance}{time}[/tex].

The distance its the length of the displacement vector,

[tex]d=\sqrt{x^2+y^2}[/tex]

[tex]d=\sqrt{(30 \ m)^2+0^2}[/tex]

[tex]d=30 \ m[/tex]

So, the average speed its:

[tex]\frac{30 \ m }{30 \ s} = 1\frac{m}{s}[/tex].

The average velocity, instead, its:

[tex]\vec{v} = \frac{displacement}{time}[/tex]

[tex]\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}[/tex]

[tex]\vec{v} = (1 \ \frac{m}{s} ,\ 0)[/tex]

This is, 1 m/s north.

Answer:

Current through each resistor is 2 A.

Explanation:

Given that,

Voltage of a battery, V = 5 volts

Resistance 1, R = 1.25 ohms

Resistance 2,R' = 1.25 ohms

Both resistors are connected in series. The equivalent resistance is given by :

R" = R + R'

R" = 1.25 + 1.25

R" = 2.5 ohms

The current flowing throughout all resistors is same in series combination of resistors. Current can be calculated using Ohm's law as :

[tex]I=\dfrac{V}{R"}[/tex]

[tex]I=\dfrac{5\ V}{2.5}[/tex]

I = 2 A

So, the current through each resistor is 2 A. Hence, this is the required solution.

Answer:

Height, h = 300.27 meters

Explanation:

Given that,

Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]

We need to find the height of the manometer required. The pressure at a height is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]

h is the height of the manometer required.

[tex]h=\dfrac{P}{\rho g}[/tex]

[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]

h = 300.27 meters

So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.

Answer:

The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]

Explanation:

For this particular situation, the angular frequency of the system is given by

[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]

Answer:

The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]

Solution:

As per the question:

Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]

Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex]        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]

[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]

[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]

Q = [tex]\pm 5.707\times 10^{13} C[/tex]

Answer:

frequency = 19.56 Hz

Explanation:

given data

length L = 7 m

mass m = 7 g

mass M = 2.50 kg

distance d = 4 m

to find out

fundamental frequency

solution

we know here frequency formula is

frequency = [tex]\frac{v}{2d}[/tex]   ...........1

so here d is given = 4

and v = [tex]\sqrt{\frac{T}{\mu} }[/tex]  ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

and μ = [tex]\frac{m}{l}[/tex] =  [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m

so from equation 2

v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]

v = 156.52

and from equation 1

frequency = [tex]\frac{v}{2d}[/tex]

frequency = [tex]\frac{156.52}{2(4)}[/tex]

frequency = 19.56 Hz

Final answer:

The fundamental frequency of the vibrating cord is approximately 14.18 Hz.

Explanation:

To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:

f1 = 1/2L * sqrt(T / μ)

Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.

Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)

Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz

Answer:

The angle is [tex]\theta\approx 14.61[/tex] degrees.

Explanation:

Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:

[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]

If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.

[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].

If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].

[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]

The angle is thus

[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.

Answer:

[tex]F_{net}=N\ sin\theta[/tex]

Explanation:

Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.    

The attached free body diagram shows the car on the banked turn. It is clear that,

In vertical direction,

[tex]N\ cos\theta=mg[/tex]

In horizontal direction,

[tex]F_{net}=F_{centripetal}[/tex]

[tex]F_{net}=N\ sin\theta[/tex]

So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Where, [tex]\mu[/tex] = mass per unit length

T = tension

Put the value into the formula

[tex]v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}[/tex]

[tex]v =519.61\ m/s[/tex]

Hence, The speed of transverse waves in this string is 519.61 m/s.

Answer:

(a) [tex]E_{ c} = 112.5 \mu J[/tex]

(b) [tex]E'_{ c} = 18 \mu J[/tex]

Solution:

According to the question:

Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]

Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]

(a)The Energy stored in the Capacitor is given by:

[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]

[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]

[tex]E_{ c} = 112.5 \mu J[/tex]

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]

[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]

[tex]E'_{ c} = 18 \mu J[/tex]

(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]

(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]

The given parameters;

The energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]

When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]

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Answer:

Explanation:

mass of astronaut, m = 60 kg

mass of space craft, M = 120 kg

t = 1.5 s

Force on space craft = 30 N

(a) According to Newton's third law

Force on spacecraft by the astronaut = Force on astronaut by the space craft

Force on astronaut by the space craft  = 30 N

(b) According to Newton's second law

Force  = mass x acceleration

Let a be the acceleration of the astronaut

30 = 60 x a

a = 0.5 m/s^2

Let A be the acceleration of the spacecraft

30 = 120 x A

a = 0.25 m/s^2

Answer:26.96 m,2.34 s

Explanation:

Given

Wrench hit the ground with a speed of 23 m/s

Applying equation of motion

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

Here u=0 because it is dropped from a height of S m

[tex]23^2-0=2\times 9.81\times s[/tex]

[tex]s=\frac{529}{2\times 9.81}=26.96 m[/tex]

Time required by wrench to hit the ground

v=u+at

[tex]23=9.81\times t[/tex]

[tex]t=\frac{23}{9.81}=2.34 s[/tex]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance =  3.6 m

v = Image distance

f = Focal length = 35 mm

[tex]h_u[/tex]= Object height = 1.8 m

a) Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm[/tex]

The CCD sensor is 35.34 mm from the lens

b) Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm[/tex]

The person appears 17.67 mm tall on the sensor

Answer:48 %

Explanation:

Given

outside temperature [tex]=25^{\circ}\approx 298 K[/tex]

Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]

Maximum efficiency is carnot efficiency which is given by

[tex]\eta =1-\frac{T_L}{T_H}[/tex]

[tex]\eta =1-\frac{298}{573}[/tex]

[tex]\eta =\frac{275}{573}=0.4799[/tex]

i.e. [tex]47.99 %\approx 48 %[/tex]

No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible

Answer:

Explanation:

Capacity of a parallel plate capacitor  C = ε₀ A/ d

ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.

C =(  8.85 X10⁻¹² X  27 X 10⁻⁴ ) / 3 X 10⁻³

= 79.65 X 10⁻¹³ F.

potential diff between plate = Charge / capacity

= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³

= 601 V

Electric field = V/d

= 601 / 3 x 10⁻³

= 2 x 10⁵ N/C

Force on proton

= charge x electric field

1.6 x 10⁻¹⁹ x 2 x 10⁵

= 3.2 x 10⁻¹⁴

Acceleration a = force / mass

= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷

= 1.9 x 10¹³ m s⁻²

Distance travelled by proton = 3 x 10⁻³

3 x 10⁻³ = 1/2 a t²

t = [tex]\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }[/tex]

t = 1.77 x 10⁻⁸ s

Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.

Plugging the values into the equation:
 vi = d/t
 vi = 1.19 m / 0.20 s
 vi = 5.95 m/s

Therefore, the initial velocity of the ball is approximately 5.95 m/s.

Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.

To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:

t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s

Therefore, the time it takes to reach the peak is approximately 0.61 seconds.

As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:

Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s

Hence, it should take approximately 1.22 seconds before the ball reappears.

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Answer:

93750 ft/s²

Explanation:

t = Time taken

u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)

v = Final velocity = 0

s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]

a = Acceleration

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]

The magnitude of acceleration is 93750 ft/s²

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

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

By substituting the values

[tex]1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}[/tex]

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, [tex]v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )[/tex]

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

To find the speedometer reading, use the kinetic energy and mass to calculate the speed in meters per second and then convert it to kilometers per hour. The calculated speed is approximately 88 km/hr.

To determine the speedometer reading of the car in km/hr based on the kinetic energy given, we use the kinetic energy formula KE = ½ mv². We rearrange this formula to solve for speed (v): v = √(2KE/m).

Plugging in the provided kinetic energy of 1.2 x 10´ J and the car's mass of 1000 kg, we get:

v = √(2 × 1.2 x 10´ J / 1000 kg)

v = √(24 x 10´ J/kg)

v ≈ 24.49 m/s

To convert from m/s to km/hr, we multiply by 3.6:

v = 24.49 m/s × 3.6 km/hr per m/s ≈ 88.16 km/hr

Therefore, the speedometer would read approximately 88 km/hr.

Answer:

Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].

Explanation:

Assuming,

Given:

The density of an object is defined as the mass of the object per unit volume.

Therefore,

[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]

Since only 95% of the body of the swimmer is inside the water, therefore,

[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]

According to Archimedes' principle,

[tex]m=m_w\\[/tex]

Using (1),

[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]

Answer:

For 0.24 sec the person was in the air.

Explanation:

Given that,

Height = 1 m

Initial velocity = 3 m/s

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Where, u = initial velocity

s = height

Put the value into the formula

[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]4.9t^2+3t-1=0[/tex]

[tex]t = 0.24\ sec[/tex]

On neglecting the negative value of time

Hence, For 0.24 sec the person was in the air.

Explanation:

Given that,

Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]

Distance between charges, r = 1.81 m

(a) The electrostatic force that one charge exerts on the another is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]

[tex]F=9.21\times 10^{-8}\ N[/tex]

(b) As both charges are positively charged. So, the force of attraction between them is repulsive.

Answer:

Electric field at distance of 50 micrometer due to a proton is 0.576 N/C

Explanation:

We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]

We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]

Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]

Answer:

Statement E is false.

Explanation:

Statement a says that the object would have moved a distance of 10 m.

At t = 0

x0  = 0

y0 = 0

At t= 2

x2 = 2 * 2^2 = 8

y2 = 3 * 2 = 6

[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]

[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]

Statement A is true.

Statement B says that the momentum of the object would be 4*i+1.5*j

To test this we need to know the speed.

The speed is the first derivative by time:

v = 4*t*i + 3*j

At t = 2 the speed is

v2 = 8*i + 3*j

The momentum is

P = m * v

P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)

This statement is true.

Statement C says that the kinetic energy of the object would be 18 J.

The magnitude of the speed is

[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]

The kinetic energy is

Ek = 1/2 * m * v^2

Ek = 1/2 * 0.5 * 8.54^2 = 18 J

Statement C is true

Statement D says that the force acts perpenducular to the direction of the unit vector j.

The acceleration is the second derivative respect of time

a = 4*i + 0*j

Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.

Statement D is true.

Statement E says that the force acting on the object is of 4 N

f = m * a

f = 0.5 * 4 = 2 N

Statement E is false.

Answer:

(C) a circle starting at time t=0 on the positive x axis

Explanation:

particle's position is

r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^

this is a parametric equation of a circle, because the axis at x and y are the same = R.

for t=0:

r=Ri^

so: circle starting at time t=0 on the positive x axis

On the other hand:

[tex]v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j][/tex]

The value of the magnitude of the acceleration is:

[tex]a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}[/tex]

we can recognise that this represent the centripetal acceleration.

Answer:

The life time of the particle is [tex]2.491\times 10^{- 25} s[/tex]

Solution:

As per the question:

Average rest energy of [tex]Z^{0}boson = 91.19 GeV[/tex]

Uncertainty in rest energy, [tex]\Delta E_{r} = 2.5 GeV = 2.5\times 10^{9}\times 1.6\times 140^{- 19} J = 4\times 10^{- 10} J[/tex]

Now,

From the Heisenberg's Uncertainty Principle, we can write:

[tex]\Delta E_{r}\times \Delta T \geq \frac{h}{2\pi}[/tex]

where

T = Life time  of the particle

[tex]\Delta T \geq \frac{h}{2\pi\Delta E_{r}}[/tex]

[tex]\Delta T \geq \frac{6.262\times 10^{- 34}}{2\pi\times 4\times 10^{- 10}}[/tex]

[tex]\Delta T \simeq 2.491\times 10^{- 25} s[/tex]

The lifetime of the Z0 boson is approximately 8.95 x 10^-17 seconds.

The Z0 boson is a particle that mediates the weak nuclear force. Its average rest energy is 91.19 GeV and it has an intrinsic width of 2.5 GeV. The lifetime of a particle can be calculated using the Heisenberg uncertainty principle, which relates the energy uncertainty to the time uncertainty. The relationship is given by the equation ΔE × Δt ≥ ℏ/2, where ℏ is the reduced Planck constant. By rearranging the equation and substituting the values, we can calculate the lifetime of the Z0 boson to be approximately 8.95 x 10^-17 seconds.

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Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

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

here Final Velocity v=0

a=acceleration due to gravity

[tex]0-u^2=2\left ( -g\right )h[/tex]

[tex]u=\sqrt{2gh}[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]

[tex]u=\sqrt{24.328}[/tex]

u=4.93 m/s