Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving in a circular path perpendicular to a magnetic field of magnitude B. (Use any variable or symbol stated above along with the following as necessary: q and c.)

Answer:

6.455 m/s

Explanation:

Given;

Frequency of train whistle, f = 496 Hz

Frequency observed, f' = 478 Hz

Speed of sound in air = 343 m/s

Now,

From the instant Doppler's effect, we have the relation

[tex]f = \frac{v+v_s}{v-v_s}\times f'[/tex]

here,

v is the speed of the sound

[tex]v_s[/tex] is the speed of the train

[tex]496 = \frac{343+v_s}{343-v_s}\times478[/tex]

or

1.037 × ( 343 - [tex]v_s[/tex] ) = 343 + [tex]v_s[/tex]

or

355.91 - [tex]v_s[/tex] = 343 + [tex]v_s[/tex]

or

12.91 = [tex]2v_s[/tex]

or

[tex]v_s[/tex]  = 6.455 m/s

Answer:

a)40100m/s

b)-4.348x10^- m/s^2

Explanation:

to calculate the change in the planet's velocity we have to rest the speeds

ΔV=-22.8-17.3=-40.1km/s=40100m/s

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s

t=1.71years=53926560*1.71=92214417.6

then we can use the ecuation number 1 to calculate the aceleration

Vf=-22.8km/s

Vo=17.3km/s

Vf=Vo+at

a=(vf-vo)/t

a=(-22.8-17.3)/92214417.6

a=-4.348x10^-7 km/s^2=-4.348x10^- m/s^2

Final answer:

The total change in velocity is -40,100 m/s, and the average acceleration is calculated using the formula a = (final velocity - initial velocity) / time.

Explanation:

The total change in the planet's velocity:

To find the total change in velocity, we subtract the initial velocity from the final velocity: -22.8 km/s - 17.3 km/s = -40.1 km/s. Converting this to m/s, we have -40,100 m/s.

The average acceleration of the planet:

Acceleration is the rate of change of velocity over time. Using the given time interval, we can calculate the average acceleration: a = (final velocity - initial velocity) / time. Plugging in the values, we get a = (-40,100 m/s - 17,300 m/s) / 1.71 yrs.

Answer:

Explanation:

Displacement = minimum distance between initial and final point

Here displacement = Hypotenuse AC of triangle ABC

= [tex]\sqrt{464^2+831^2}[/tex]

= 951.76 km

Direction of AC

If θ be the angle AC makes with east

Tan θ = 831 / 464 = 1.79

θ = 60.8°

South of east.

c ) average velocity= Displacement / time

Total time =( 50/60 + 1.2 ) hour

2.0333 hour

average velocity = 951.76 km / 2.0333 hour

= 468 km / h

Direction of velocity will be same as the direction of displacement.

e ) speed = distance / time

distance travelled

= 464 + 831 = 1295 km

speed = 1295 / 2.0333

= 636.9 km/ h .

Answer:

The force of weight and drag force of the air

Explanation:

This is because when the ball leaves the hand the only force acting on it is the gravity of the earth, therefore its weight, since the force applied by the quarterback is no longer being applied to the ball. You can also consider the drag force of the air in the ball but in some cases it is minimal and can be neglected.

The forces acting on the ball as it leaves the hand of the quarterback are;

The upward motion of an object thrown upwards is reduced by the opposing force. These opposing forces act in opposite direction to the motion of the object.

When the football is thrown upward, the force from the quarterback provides the thrust force with the ball moves upwards, after the ball leaves the hand of the quarterback the upward motion is reduced by opposing forces.

These opposing forces include the following;

Thus, form the given options, the forces acting on the ball as it leaves the hand of the quarterback are the drag force from the air and the weight force from the Earth.

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Answer:

Part a)

[tex]\lambda = 4\pi[/tex]

Part b)

[tex]f = 47.7 Hz[/tex]

Part c)

[tex]v = 600 m/s[/tex]

Explanation:

Part a)

As we know that angular wave number is given as

[tex]k = \frac{2\pi}{\lambda}[/tex]

[tex]k = 0.500[/tex]

[tex]0.500 = \frac{2\pi}{\lambda}[/tex]

[tex]\lambda = \frac{2\pi}{0.500}[/tex]

[tex]\lambda = 4\pi[/tex]

Part b)

As we know that angular frequency is given as

[tex]\omega = 300 rad/s[/tex]

[tex]\omega = 2\pi f[/tex]

[tex]300 = 2\pi f[/tex]

[tex]f = \frac{300}{2\pi}[/tex]

[tex]f = 47.7 Hz[/tex]

Part c)

Speed of the wave is given as

[tex]v = \lambda \times frequency[/tex]

so we have

[tex]v = 4\pi \times 47.7[/tex]

[tex]v = 600 m/s[/tex]

Answer:

The train should be a 10 meters away before the professor falls

Explanation:

Calculation of the teacher's fall time:

[tex]y(t)=v_{o}t-1/2*g*t^{2}[/tex]

in this case Vo=0 and y=-5:

[tex]-5=-1/2*10*t^{2}[/tex]

[tex]t=1s[/tex]

Calculation how far away is the train when the professor falls:

The distance the train travel while the professor is falling:

[tex]d=v_{train}*t=10m/s*1s=10m[/tex]

So the train should be a 10 meters away before the professor falls

Answer:

Doubles

Explanation:

Before starting to fall, the object will have energy in the form of potential energy. This will be the energy that will be transformed into kinetic energy, once the object starts its fall. And the potential energy that the object had before starting to fall, will be equal to the kinetic energy acquired during the fall. The potential energy is given by:

[tex]E_p = m*g*h[/tex]

Where m is the mass of the object, g is the gravity, and h is the distance to the floor. As you can see, The potential energy is proportional to the height of the fall, so if this height doubles, the potential energy and therefore, the kinetic energy acquired will doubles.

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that

[tex]v_f = 0[/tex]

also we have

[tex]a = -9.81 m/s^2[/tex]

so we have

[tex]0 - v_i^2 = 2(-9.81)(16)[/tex]

[tex]v_i = 17.72 m/s[/tex]

Now we need to find the height where its speed becomes half of initial value

so we have

[tex]v_f = 0.5 v_i[/tex]

now we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex](0.5v_i)^2 - v_i^2 = 2(-9.81)h[/tex]

[tex]-0.75v_i^2 = -19.62 h[/tex]

[tex]0.75(17.72)^2 = 19.62 h[/tex]

[tex]h = 12 m[/tex]

Answer:

If an object moves twice as fast its kinetic energy quadruples.

Explanation:

The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:

K₁= 1/2 * m* v²

If we double the speed of the body, its kinetic energy (K₂) will be:

K₂= 1/2 * m*( 2v)²

K₂= 1/2 * m* 4 *v²

K₂= 4(1/2 * m *v²)

K₂= 4*K₁

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

[tex]\omega=-\alpha t[/tex]

[tex]\alpha=-\dfrac{\omega}{t}[/tex]

[tex]\alpha=-\dfrac{50.0}{20.0}[/tex]

[tex]\alpha=-2.5\ rad/s^2[/tex]

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t[/tex]

Put the value into the formula

[tex]\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2[/tex]

[tex]\theta=500\ rad[/tex]

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

[tex]\vec{\tau}=\vec{r}\times\vec{f}[/tex]

[tex]\tau=r\times f\sin\theta[/tex]

Put the value into the formula

[tex]\tau=2.5\times4\times 9.8\sin60[/tex]

[tex]\tau=84.87\ N-m[/tex]

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Answer:[tex]-18.05 m/s^2[/tex]

Explanation:

Given

Initial velocity of rocket A ([tex]v_A[/tex]) 5300m/s

Initial velocity  of rocket B([tex]v_B[/tex]) 8700 m/s

after time t they have displacement=0

[tex]s_A=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]

[tex]0=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]

[tex]t=\frac{5300\times 2}{11}=963.63 s[/tex]

for Rocket B

[tex]s_B=8700\times t+\frac{1}{2}\times \left ( a_B\right )t^2[/tex]

time is same for A & B

[tex]0=8700\times 963.63+\frac{1}{2}\times \left ( a_B\right )963.63^2[/tex]

[tex]a_B=\frac{2\times 8700}{963.63}=-18.05 m/s^2[/tex]

Answer:

0.235 nC

Explanation:

Given:

Since the electric field is the electric force applied on a charged body of unit charge.

[tex]\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC[/tex]

Hence, the value of q is 0.235 nC.

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east  is = x

so distance from starting point = 5.21 × 10³ - x        ...................1

and we can say time required to reach end

time required = distance / speed

time required = [tex]\frac{5.21 *10^3 - x}{1.32}[/tex]    ................2

and

time required for 6.44 km west

time required = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]    ................3

and time required for distance x

time required = [tex]\frac{x}{0.687}[/tex]    ................4

so from equation 2 , 3 and 4

[tex]\frac{5.21 *10^3 - x}{1.32}[/tex] = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex]  + [tex]\frac{x}{0.687}[/tex]

x = 1273.78 m

so distance in east is 1273.78 m

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at²    .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m

Answer:

a) 2.24°

b) 11.99 km

c) 0.9992 miles

Explanation:

We can think of the road as a triangle with a 12 km hypotenuse and a 0.5 km side. Then:

l = 12

h = 0.5

and

sin(a) = h / l

a = arcsin(h / l)

a = arcsin(0.5 / 12)

a = 2.24°

That is the slope.

The map distance travelled would be the other side of the triange.

cos(a) = d / l

d = l* cos(a)

d = 12 * cos(2.24) = 11.99 km

And for miles

d = 1 mile * cos(2.24) = 0.9992 miles

Answer:

[tex]\rm 1.374\times 10^6\ N/C.[/tex]

Explanation:

Given:

The surface area of each of the disc is given by

[tex]\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.[/tex]

For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by

[tex]\rm E=\dfrac{|\sigma|}{2\epsilon_o}[/tex]

where,

The electric field strength between the discs due to negative disc is given by

[tex]\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]

Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.

The electric field strength between the discs due to positive disc is given by

[tex]\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]

The direction of this electric field is away from the positive disc, i.e., towards negative disc.

The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by

[tex]\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.[/tex]

Answer: 84.63 m

Explanation:

The speed of sound in the air, at [tex]0 \°C[/tex] is [tex]331.5 m/s[/tex], and for each degree Celsius the temperature rises, the speed of the sound increases by [tex]0.6 m/s[/tex]. So, if we estimate the speed of sound at [tex]30 \°C[/tex] it will be [tex]349 m/s[/tex].

On the other hand, the speed of sound [tex]V[/tex] is defined as the distance traveled [tex]d[/tex] in a especific time [tex]t[/tex]:  

[tex]V=\frac{d}{t}[/tex]  

Where:  

[tex]V=349 m/s[/tex] is the speed of sound  

[tex]t=\frac{0.485 s}{2}=0.2425 s[/tex] is half the time the sound wave travels since you say "hello", the sound wave hits the building and then returns to you

[tex]d[/tex] is the distance between you and the building

[tex]349 m/s=\frac{d}{0.2425 s}[/tex]  

Finding [tex]d[/tex]:  

[tex]d=(349 m/s)(0.2425 s)[/tex]  

Finally:  

[tex]d=84.63 m[/tex]

To find the distance to the building, calculate the speed of sound at 30.6°C, multiply it by the echo time of 0.485 seconds, and divide by 2 to account for the round trip of the sound. Then, convert the distance from meters to feet to get approximately 280.95 feet.

To calculate the distance of the building when an echo is heard 0.485 seconds after shouting, we need to consider the speed of sound at the given air temperature of 30.6°C. Since the speed of sound varies with temperature, we use the formula v = v_0 + 0.6t, where v_0 is the speed of sound at 0°C (approximately 331.5 m/s), t is the temperature in Celsius, and v is the speed of sound at t degrees Celsius. At 30.6°C, the speed of sound is v = 331.5 m/s + 0.6(30.6°C), which calculates to roughly 349.86 m/s. Since the echo travels the distance to the building and back, the actual distance to the building is half the total distance the sound wave traveled.

Therefore, the one-way distance to the building is 0.485 s × 349.86 m/s ÷ 2. Converting meters to feet (1 meter = 3.28084 feet), we find that the building is approximately 280.95 feet away.

Answer:

Part A)   x=40m at t= 2.5s

Part B)    x=240m at t=15 s

Explanation:

Data:

x₁=0 , t₁=0

x₂=80m,  t₂=5 s

t₃= 2.5 s

t₄= 15 s

problem development

We set the ratios x / t:

Part A)

[tex]\frac{x_{3} }{t_{3} } =\frac{x_{2} }{t_{2} }[/tex]

[tex]\frac{x_{3} }{2.5} =\frac{80}{5 }[/tex]

x₃=16*2.5

x₃=40m

Part B)

[tex]\frac{x_{4} }{t_{4} } =\frac{80 }{5 }[/tex]

[tex]\frac{x_{4} }{15} =\frac{80}{5 }[/tex]

x₄=16*15

x₄=240m

Final answer:

The car's position at t = 2.5 s will be 40 m as it is half of the interval to reach 80 m. At t = 15 s, three times the interval, the car will be at 240 m assuming constant velocity.

Explanation:

Given that the car moves from x1 = 0 m at t1 = 0 s to x2 = 80 m at t2 = 5.0 s, we can determine its position at other times under the assumption of constant velocity. Because the velocity is constant, the ratios of times to positions are constant as well.

Part A: Car's Position at t = 2.5 s

The time t = 2.5 s is exactly half of the time interval given (5.0 s), so the position should be half of 80 m, which is 40 m.

Part B: Car's Position at t = 15 s

To find the position at t = 15 s, let's first determine the time ratio. The ratio of 15 s to 5.0 s is 3:1. Since the velocity is constant, the positions scale with time. So, the position x at t = 15 s is three times 80 m, giving us 240 m.

Answer:

The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Solution:

As per the question:

[tex]v_{1} = 3 x 10^5 m/s[/tex]

[tex]v_{2} = 6.40 x 10^6 m/s[/tex]  

length of the re region, l = 1 cm = 0.01 m

Now,

Using the third equation of motion:

[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]  

[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]  

[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]  

[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]  

Answer:

The charge at the center of the conducting sphere is zero.

Explanation:

A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.

Answer:

9 m/s

Explanation:

mass of cannon, M = 100 kg

mass of cannon ball, m = 10 kg

velocity of cannon ball, v = 90 m/s

Let the recoil velocity of cannon is V.

Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.

Momentum before the firing = momentum after the firing

M x 0 + m x 0 = M x V + m x v

0 = 100 x V + 10 x 90

V = - 9 m/s

Thus, the recoil velocity of cannon is 9 m/s.

Answer:

40 m

Explanation:

A car travelling at 45 ms has a kinetic energy of

Ec = 1/2 * m * v^2

We do not know the mass, so we can use specific kinetic energy:

ec = 1/2 * v^2

In this case

ec = 1/2 * 45^2 = 1012 J/kg

If it stops in 10 m, the braking force performed a sppecific work of 1012 J/kg in 10 m

L = F * d

F = L / d

F = 1012 / 10 = 101.2 N/kg

If the car is running at 90 m/s, the specific kinetic energy of:

ec2 = 1/2 * 90^2 = 4050 J/kg

With the same braking force the braking distance is:

d2 = L2 / F

d2 = 4050 / 101.2 = 40 m

Answer:

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

Explanation:

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

Force between them F = 8 X 10⁻³ N distance between them d = 4.1 m

F = k ( q₁ xq₂ ) / d²

8 x 10⁻³ = [tex]\frac{8.99\times10^9\times q_2\times (10^{-5}-q_2)}{(4.1)^2}[/tex]

14.95 x 10⁻¹² = 10⁻⁵q₂ - q₂²

q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹² =0

This is a quadratic equation .The solution of it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

q₁ = 3.2 x 10⁻⁶ C

q₂ = 6.8 x 10⁻⁶C

Charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.

It is also known as electric charge, electrical charge, or electrostatic charge and symbolized q.

Given,

First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )

F = 8 X 10⁻³ N

d = 4.1 m

By using the Formula,

[tex]F =\frac{k ( q_{1} * q_{2} ) }{d^{2} }[/tex]

8 x 10⁻³ = [tex]\frac{8.99*10^{9}*q_{2} *10^{-5} -q_{2} }{(4.1)^{2} }[/tex]

14.95 x 10⁻¹²

= 10⁻⁵q₂ - q₂²

= q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹²=0

Therefore,

The above equation is a quadratic equation , it gives the value of q₂

q₂ = 6.8 x 10⁻⁶C

q₁ = 3.2 x 10⁻⁶ C

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Answer:

115 m

Explanation:

The coin is in free fall, subject only to the acceleration of gravity. We set up a frame of reference with the 0 at ground and the X axis pointing upwards. We use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 will be the height of the building as it is the starting point of the coin.

We have one equation with 2 unknowns (V0 and X0), we need another equation.

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

We know that at t = 1.29 s it will  reach the highest point, at this point speed is zero.

V0 = V(t) - a * t

V0 = 0 - (-9.81) * 1.29 = 12.65 m/s.

Now we can go back to the other equation:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = X(t) - V0 * t - 1/2 * a * t^2

X0 = 0 - 12.65 * 6.3 - 1/2 * (-9.81) * 6.3^2 = 115 m

Answer:

The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).

If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

Answer:

Given, average lifetime of muons at rest is To = 2.21μs

Average lifetime of moving muons, T = 7.04μs

mass of muon = 207 mass of electron

a)

rest mass energy of an electron is 0.511 MeV.

[tex] T =\frac{T_o}{\sqrt{1-\beta^2}}[/tex]

Substitute the values:

[tex]7.04 =\frac{2.21}{\sqrt{1-\beta^2}}\\ \beta^2 = 0.90146\\ \beta = 0.95[/tex]

b) K = (γ-1) m₀c²

[tex]\gamma = \frac{1}{\sqrt{1-\beta^2}} = 3.185[/tex]

K = (3.185-1) (207 ×9.1×10⁻³¹) (3×10⁸)² = 3.7×10⁻¹¹J

c) p = γmv = (3.185)((207 ×9.1×10⁻³¹)(0.95×3×10⁸)= 1.71×10⁻¹⁹ N/s

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required [tex]= \frac{(14.644 x 10^6)}{784}[/tex]

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

Answer:

option D ) is correct.

Explanation:

When two charges Q₁ and Q₂ which are unequal in magnitude as well as in nature are at a distance of R then a force F is created between them which is given by the expression

 F = k ( Q₁ X Q₂) / R²

This force acts equally on both the charges Q₁ and Q₂ irrespective of the fact that they are unequal charges . In other words both Q₁ and Q₂ experience same force F whose value is given above.

In the given case too charges are unequal in magnitude and in nature . They will attract each other with equal force . So if A experiences force F , B too will experience the same force,

Hence option D ) is correct.

Answer:

Truck will be at a distance of 720 m

Explanation:

Speed of the truck is given as 90 m/sec

Time for which truck travel t = 10 seconds

We have to find the distance that how much truck travel in 8 sec with speed of 90 m/sec

We know that distance is given by [tex]distance=velocity\times time[/tex]

So distance traveled by truck in 8 sec will be equal to [tex]=90\times 8 =720m[/tex]

So truck will be at a distance of 720 m