A free positive charge released in an electric field will:
accelerate in the direction opposite the electric field.
accelerate along a circular path.
accelerate in the direction in which the electric field is pointing.
remain at rest.
accelerate in a direction perpendicular to the electric field.

Final answer:

To find the initial speed of a ball thrown to a maximum height of 36 m, we use kinematic equations that factor in the acceleration due to gravity. The ball's initial speed can be calculated using the formula for objects under constant acceleration, considering that the final velocity at the max height is 0 m/s.

Explanation:

Calculating the Launch Speed of the Ball

To determine the initial speed at which the ball was thrown to reach a maximum height of 36 m, we can use the principles of kinematics under the influence of gravity. In the absence of air resistance, a ball thrown upwards will decelerate at a rate equal to the acceleration due to gravity until it comes to a stop at its maximum height. We use the following kinematic equation for an object under constant acceleration:

s = ut + 1/2at^2

Where:

s is the displacement (maximum height in this case, which is 36 m)

u is the initial velocity (what we want to find)

a is the acceleration due to gravity (-9.81 m/s^2, the negative sign indicates acceleration is in the direction opposite to the initial velocity)

t is the time taken to reach the maximum height (not needed in this calculation)

At the maximum height, the final velocity (v) is 0 m/s, so we use the following equation:

v^2 = u^2 + 2as

Plugging in the known values:

0 = u^2 + 2(-9.81 m/s^2)(36 m)

u^2 = 2(9.81 m/s^2)(36 m)

u = √(2(9.81 m/s^2)(36 m))

The initial speed u can be calculated from this equation to find out with what speed the ball was thrown to achieve a 36 m height.

Answer:

The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].

Explanation:

Given that,

Charge on [tex]q_{1}=-6.0\ \mu C[/tex]

Distance [tex]x= -3.0\ m[/tex]

Charge on [tex]q_{2}=1.0\ \mu C[/tex] at origin

Distance  [tex]x= 3.0\ m[/tex]

Charge on [tex]q_{3}=-1.0\ \mu C[/tex]

We need to calculate the electric force on q₁

Using formula of electric force

[tex]F_{12}=\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F_{12}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times1.0\times10^{-6}}{(3.0)^2}[/tex]

[tex]F_{12}=-0.006\ N[/tex]

Negative sign shows the attraction force.

We need to calculate the electric force F₁₃

[tex]F_{13}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times(-1.0\times10^{-6})}{(6.0)^2}[/tex]

[tex]F_{13}=0.0015\ N[/tex]

Positive sign shows the repulsive force.

We need to calculate the net electric force

[tex]F=F_{12}+F_{13}[/tex]

[tex]F=-0.006+0.0015[/tex]

[tex]F=0.0045\ N[/tex]

[tex]F=4.5\times10^{-3}\ N[/tex]

Hence, The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].

The correct answer is that the electric force on q1 is 2.55 x 10^5 N, directed towards the origin.

To find the electric force on q1, we need to calculate the net electric force due to q2 and q3. According to Coulomb's law, the force between two point charges is given by:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

[tex]where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \) N m^2/C^2), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.[/tex]

First, we calculate the force between q1 and q2:

[tex]\[ F_{12} = k \frac{|q_1 q_2|}{r_{12}^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m})^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(9.0 \text{ m}^2)} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{9.0 \text{ m}^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (6.67 \times 10^{-13} \text{ C}^2/\text{m}^2) \][/tex]

[tex]\[ F_{12} = 5.99 \times 10^{-3} \text{ N} \][/tex]

Since q1 and q2 have opposite signs, the force [tex]\( F_{12} \)[/tex] is attractive, meaning it is directed towards q2, which is towards the origin.

Next, we calculate the force between q1 and q3:

[tex]\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(-1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m} - 3.0 \text{ m})^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(6.0 \text{ m})^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{36.0 \text{ m}^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (1.67 \times 10^{-12} \text{ C}^2/\text{m}^2) \][/tex]

[tex]\[ F_{13} = 1.50 \times 10^{-2} \text{ N} \][/tex]

Since q1 and q3 have the same sign, the force [tex]\( F_{13} \)[/tex] is repulsive, meaning it is directed away from q3, which is also towards the origin in this case.

Now, we add the forces vectorially. Since both forces are directed towards the origin, we can simply add their magnitudes:

[tex]\[ F_{net} = F_{12} + F_{13} \][/tex]

[tex]\[ F_{net} = 5.99 \times 10^{-3} \text{ N} + 1.50 \times 10^{-2} \text{ N} \][/tex]

[tex]\[ F_{net} = 2.099 \times 10^{-2} \text{ N} \][/tex]

Converting this to a more standard scientific notation:

[tex]\[ F_{net} = 2.55 \times 10^5 \text{ N} \][/tex]

Therefore, the electric force on q1 is [tex]\( 2.55 \times 10^5 \)[/tex] N, directed towards the origin.

Answer:

The runners are 0.159 mi to the west of the flagpole.

Explanation:

Let´s place the origin of the frame of reference at the point where the flagpole is located.

The initial position of runner A is 3.91 mi and his velocity is -6.04 mi/h

The initial position of runner B is -2.95 mi and his velocity is  5.00 mi/h

When both runners meet, their position is the same. The equation of the position of each runner is:

x = x0 + v · t

Where

x = position at time t

x0 = initial position

v = velocity

t = time

Then, at the meeting point:

x runner A = x runner B

3.91 mi - 6.04 mi/h · t = -2.95 mi + 5.00m/h · t

Solving for t:

3.91 mi + 2.95 mi =  5.00m/h · t + 6.04 mi/h · t

6.86 mi = 11.04 mi/h · t

t = 0.621 h

Now, we use this time to find the meeting point. We can use the equation of any runner. Let´s use the position of runner A:

x = 3.91 mi - 6.04 mi/h · 0.621 h = 0.159 mi

Since the position is positive, the runners met 0.159 mi to the west of the flagpole.

Answer:

b) 58 kg

Explanation:

Gravitational potential energy is the energy that an object has due to its state or position in which it rests.

Gravitational potential energy = U = mass x gravity x height = m g h = 474 J

Height of the stool = h = 0.84 m

rearranging m g h  and solving for the mass m gives m =

474 / (9.8)(0.84)  = 57.6 kg = 58 kg (rounded to 2 significant digits).

Final answer:

The speed of the box as it leaves the spring can be calculated using conservation of energy, accounting for the initial potential energy in the spring and work done by friction. The maximum speed of the box is achieved when all the spring's potential energy is converted into kinetic energy, before friction does any work.

Explanation:

To solve for the speed of the box at the instant it leaves the spring and its maximum speed during the motion, we can apply the principle of conservation of energy and the work-energy theorem.

Part (a): Speed of Box When It Leaves the Spring

The initial potential energy stored in the compressed spring is equal to the kinetic energy of the box plus the work done by friction as the box slides the distance where the spring is compressed:

PE_spring = KE_box + Work_friction

Using the formula for potential energy PE_spring = (1/2)[tex]kx^2[/tex], where k is the spring constant and x is the compression distance, and the work done by friction Work_friction = μ_kmgx, where μ_k is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity, we can find the kinetic energy which will then be used to calculate the speed using KE = (1/2)[tex]mv^2[/tex].

To find the actual speed, we rearrange to v = √(2(PE_spring - Work_friction)/m).

Part (b): Maximum Speed of the Box

The maximum speed of the box occurs at the point where all the potential energy in the spring has been converted to kinetic energy and before any work has been done by friction. Therefore, v_max = √(2PE_spring/m).

Answer:

power = 384.92 W

Explanation:

given data

resistance = 44.0 Ω

impedance = 71 Ω

ΔVrms = 210 V

to find out

average power

solution

we know here power formula that is

power =  [tex]\frac{V^2rms}{impedance}[/tex]cos∅ .........1

we know here cos∅  =  [tex]\frac{resistance}{impedance}[/tex]

cos ∅  =  [tex]\frac{44}{71}[/tex] = 0.6197

so from equation 1

power =  [tex]\frac{210^2}{71}[/tex] × 0.6197

power = 384.92 W

Answer:

Magnitude of vector B = 6643 m

Magnitude of vector C = 7201 m

Explanation:

Knowing that the sum of the internal angles of a triangle is 180°, we can obtain the internal angles of the triangle formed by the three displacement vectors (see the attached figure, the calculated angles are in red and the given angles in black).

The angles were calculated as follows (see figure):

angle BC = 180°- 90° - 41° - 37 ° = 12°

angle AB = 180° - 90° - 26° +41° = 105°

angle AC = 180° - 105° - 12 = 63°

Once we obtain the internal angles, we can use the sine rule:

sin a/ A = sin b/ B = sin c/ C where "A" is the side opposite to the angle "a", "B" is the side opposite to the angle "b" and "C" is the side the opposite to the angle "c".

Then:

sin 12° / A = sin 63°/ B = sin (105°) / C

sin 12° / 1550 m = sin 63° / B

B = sin 63° * (1550 m / sin 12°) = 6643 m

sin 12° /1550 m = sin 105° /C

C = sin 105° * (1550 m / sin 12°) = 7201 m

Answer:

(a). The kinetic energy stored in  the fly wheel is 46.88 MJ.

(b). The time is 1.163 hours.

Explanation:

Given that,

Radius = 1.50 m

Mass = 475 kg

Power [tex]P= 15.0 hp = 15.0\times746=11190 watt[/tex]

Rotational speed = 4000 rev/min

We need to calculate the moment of inertia

Using formula of moment of inertia

[tex]I=\dfrac{1}{2}mr^2[/tex]

Put the value into the formula

[tex]I=\dfrac{1}{2}\times475\times(1.50)^2[/tex]

[tex]I=534.375\ kg m^2[/tex]

(a). We need to calculate the kinetic energy stored in  the fly wheel

Using formula of K.E

[tex]K.E=\dfrac{1}{2}I\omega^2[/tex]

Put the value into the formula

[tex]K.E = \dfrac{1}{2}\times534.375\times(4000\times\dfrac{2\pi}{60})^2[/tex]

[tex]K.E=46880620.9\ J[/tex]

[tex]K.E =46.88\times10^{6}\ J[/tex]

[tex]K.E =46.88\ MJ[/tex]

(b). We need to calculate the length of time the car could run before the flywheel  would have to be brought backup to speed

Using formula of time

[tex]t=\dfrac{46.88\times10^{6}}{11190}[/tex]

[tex]t=4189.45\ sec[/tex]

[tex]t=1.163\ hours [/tex]

Hence, (a). The kinetic energy stored in  the fly wheel is 46.88 MJ.

(b). The time is 1.163 hours.

The kinetic energy stored in the flywheel is calculated using its moment of inertia and angular velocity. The time the car could run driven by the flywheel's energy can be determined by dividing the total energy by the rate at which the energy is supplied.

The kinetic energy stored in the flywheel (part a) can be obtained using the formula for rotational kinetic energy: K.E. = 0.5 * I * ω^2, where I is the moment of inertia of the flywheel and ω is its angular velocity. The moment of inertia for a solid disc (like our flywheel) is given by I = 0.5 * m * r^2, and ω can be found by converting the given rate of 4000 rev/min to rad/s. Inputting the given values, we can calculate the kinetic energy of the flywheel.

For part b, we first convert the horsepower of the motor to Watts (1 hp = 746 Watts). This gives us the rate at which the flywheel is supplying energy. We then divide the total energy obtained in part a by this rate to find the time the car could run.

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Answer:

a) 6.028*10^10 m/s^2

b)2.156*10^-5 s

c)14.01 m

Explanation:

Hello!

I will not consider relativistic efects since the velocity of the proton is 1% of the velocity of ligth.

In order to find the acceleration we need to calculate first the force, this is done by multiplying the electric field times the charge of the proton (1e=1.6*10^-19)

[tex]ma=F=630\times1.6\times10^{-19}N[/tex]

Since the mass of the proton is 1.6726219 × 10^-27 kilograms

The acceleration it suffers due to the electric field is:

[tex]a = 6.028 \times10^{10}m/s^{2}[/tex]

Since the proton accelerates from rest, the velocity as a function of time is given by:

[tex]v = at[/tex]

So

[tex]t=\frac{1.3*10^{6}m/s}{6.028 \times10^{10}m/s^{2}}=2.156\times10^{-5}s[/tex]

Finally, the length traveled by the proton in that interval is given by:

[tex]x(t=2.156\times10^{-5}s)=\frac{1}{2} 6.028 \times10^{10}m/s^{2}\times(2.156\times10^{-5}s)^{2}=14.01 m[/tex]

Explanation:

The relationship between the wavelength and the potential difference V is given by :

[tex]\lambda=\dfrac{h}{\sqrt{2me}}\times \dfrac{1}{\sqrt{V}}[/tex]

Putting the values of known parameters,

[tex]\lambda=\dfrac{12.28}{\sqrt{V} }\times 10^{-10}\ m[/tex]

(a) [tex]V=30\ kV=30000\ V[/tex]

[tex]\lambda=\dfrac{12.28}{\sqrt{30000} }\times 10^{-10}\ m[/tex]

[tex]\lambda=7.08\times 10^{-12}\ m[/tex]

(b) [tex]V=60\ kV=60000\ V[/tex]

[tex]\lambda=\dfrac{12.28}{\sqrt{60000} }\times 10^{-10}\ m[/tex]

[tex]\lambda=5.01\times 10^{-12}\ m[/tex]

Hence, this is the required solution.

Answer:

Explanation:

Given

maximum height=0.380 m

initial velocity=[tex]v_0[/tex]

[tex]H_{max}=\frac{v_0^2}{2g}[/tex]

[tex]0.380=\frac{v_0^2}{2\times 9.81}[/tex]

[tex]v_0=2.73 m/s[/tex]

The time of flight will be [tex]t=\frac{2v_0}{g}[/tex]

time to reach top +time to reach bottom will be same

[tex]t=\frac{2\times 2.73}{9.81}[/tex]

t=0.556 s

The initial velocity of the flea as it leaves the ground is approximately 1.95 m/s. The flea is in the air for approximately 0.40 seconds.

To find the initial velocity of the flea as it leaves the ground, we can use the equation for vertical motion: v^2 = v0^2 + 2aΔy. In this equation, v is the final velocity (which is 0 at the highest point), v0 is the initial velocity, a is the acceleration due to gravity (-9.80 m/s^2), and Δy is the change in height (0.380 m). Rearranging the equation gives us: v0^2 = 2aΔy. Plugging in the values for a and Δy and solving for v0, we get v0 = sqrt(2aΔy).

For the second part of the question, we can use the equation for the time of flight: Δt = 2v0/a. Plugging in the values for v0 and a, we get: Δt = 2 * v0 / a. Calculating the value of Δt will give us the time the flea is in the air.

Using the given values for a and Δy in the equation for v0 and evaluating the expression gives us the answer to the first part of the question: v0 = sqrt(2 * 9.80 * 0.380). To find the time of flight, we can plug in the values for v0 and a into the equation for Δt: Δt = 2 * 1.95 / 9.80. Evaluating this expression gives us the answer to the second part of the question: Δt = 0.40 s. Therefore, the initial velocity of the flea as it leaves the ground is approximately 1.95 m/s and the flea is in the air for approximately 0.40 seconds.

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Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

Using the conservation of mechanical energy principle, we relate the initial potential energy in the compressed spring to the final potential energy due to the height gained by the ice cube sliding up a slope. In doing so, we calculate how far the ice cube travels up the slope before reversing direction to be approximately 0.607 meters.

The question pertains to the concept of energy conservation involving the energies of spring compression (potential energy) and gravity (also potential energy). Here the energy initially stored in the compressed spring is used to propel the ice cube up the slope until its potential energy (due to height gained) fully consumes the initial energy from the spring.

The conservation of mechanical energy principle states the initial total energy is equal to the final total energy. Initially we have spring potential energy and finally it's all converted to gravitational potential energy. So, 1/2*k*x^2=m*g*h, where k is the spring constant, x is the spring compression, m is mass of the ice block, g is acceleration due to gravity, and h is height gained by the ice cube.

We solved for h to get h = (1/2*25*(0.100)^2) / (50*9.8) = 0.255 m. This is the vertical height gained, not the distance along the slope. To get the slope distance traveled by the cube (d), we divide the height by sin(25 degrees) to get: d = 0.255/sin(25) = 0.607 m.

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Answer:

travel time is 32.4 s

maximum speed is 45.36 m/s

Explanation:

given data

distance = 980 m

acceleration = 4.20 m/s² for first 1/4 of that distance

acceleration = -1.40 m/s² for next 3/4 of that distance

to find out

travel time through the 980 m and maximum speed

solution

we know for first 1/4 of that distance is = [tex]\frac{980}{4}[/tex] = 245 m

so  equation of motion

s = ut + 0.5 ×at²     .............1

here u is initial speed = 0 and a is acceleration an t is time

s = ut + 0.5 ×at²

245 = 0+ 0.5 ×4.20 (t)²

t = 10.80 s

so

maximum speed at 1/4 of that distance

use equation of motion

v² - u² = 2as

put here value

v² - 0 = 2(4.20)× (245)

v = 45.36 m/s

so maximum speed is 45.36 m/s

and

for 3/4 distance

use equation of motion

v = u + at

here u is here 45.36 and a is acceleration and t is time and v final speed is 0

0 = 45.36 + (-1.40) × t

t = 32.4 s

so travel time is 32.4 s

Answer:

a. x= 83.03 m at t= 2 s

b. v=346.7 m/s  at t= 4s

c. t=10.5s : maximum displacement occurs

d.  t= 7s  : maximum velocity

Explanation:

Definitions

acceleration :a(t) = dv/dt :Derived from velocity with respect to time

Velocity : V(t)=dx/dt : Derived from Displacement: with respect to time

Displacement: X(t)

Developing of problem

we have a(t) =10 (-t + 2 ) (t-5)+ 100

a(t) =(-10 t + 20) (t-5)+ 100  = -10 t²+50t+ 20t-100+ 100=--10 t²+70t

a(t)=--10 t²+70t  Equation (1)

a(t) = dv/dt

-10 t²+70t=dv/dt

dv=(-10 t²+70t)dt

We apply integrals to both sides of the equation

∫dv=∫(-10 t²+70t)dt

[tex]v=\frac{-10t^{3} }{3}   +\frac{70t^{2} }{2} +C_{1}[/tex]

[tex]v=\frac{-10t}{3} +35t^{2} +C_{1}[/tex]

at time t=0, v=0, then, C₁=0

[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] Equation (2)

[tex]v=\frac{dx}{dt}[/tex]

dx= vdt

We apply integrals to both sides of the equation and we replace v(t) of the equation (2)

[tex]\int\limits {\, dx =\int\limits{\frac{(-10t^{3} }{3}+35t^{2} ) } \, dt[/tex]

[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +C_{2}[/tex]

at time t=0, x=3,  then,C₂=3

[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +3}[/tex] Equation (3)

a)Displacement at t=2s

We replace t=2s in the  Equation (3)

[tex]x=\frac{-5(2)^{4} }{6} +\frac{35(2)^{3} }{3} +3}[/tex]

x= 83.03 m at t= 2 s

b) Velocity at  t=4s

We replace t=4s in the  Equation (2)

[tex]v=\frac{-10(4)^{3}}{3} +35(4)^{2}}[/tex]

v=346.7 m/s  at t= 4s

c)Time at which maximum displacement occurs

at maximum displacement :v= [tex]\frac{dx}{dt} =0[/tex]  

in the Equation (2) :

[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] =0

[tex]35t^{2} } =\frac{10t}{3}[/tex]  ,we divide by t on both sides of the equation

[tex]t=\frac{35*3}{10}[/tex]

t=10.5s

d.  maximum velocity over the interval

at maximum velocity :a= [tex]\frac{dv}{dt} =0[/tex]  

in the Equation (1)

-10 t²+70t = 0   we multiply the equation by -1 and factor

10t ( t-7) =0

t= 7 s

Answer:

Vertical circle.

Explanation:

According to the question:

When the stone tied string moves in horizontal circle, then the tension in the string is provided by the centripetal force.

When the stone tied string moves in a vertical circle, then the tension is provided by the centripetal force as well as its weight.

Thus the probability of breaking in the vertical circle is more.

Refer to the fig shown:

In case of horizontal circle:

The centripetal force for a circle of radius R is given by:

[tex]F_{c} = \frac{mv_{P}^{2}}{R} = mg[/tex]

Now, at point P:

[tex]mg - T = \frac{mv_{P}^{2}}{R} [/tex]

Since, T = 0

[tex]mg - 0 = \frac{mv_{P}^{2}}{R} [/tex]

[tex]v_{P}^{2} = Rg[/tex]                              (1)

Now, at point Q:

[tex]T - mg = \frac{mv_{Q}^{2}}{R} [/tex]        (2)

Also, by using the law of conservation of energy, total mechanical energy at point P and Q will be conserved:

[tex]\frac{1}{2}mv_{P}^{2} + mg(2R) = \frac{1}{2}mv_{Q}^{2} + mg(0)[/tex]

Using eqn (1):

[tex]\frac{1}{2}gR + 2gR = \frac{1}{2}v_{Q}^{2}[/tex]

[tex]v_{Q}^{2} = 5gR[/tex]                                (3)

Now, using eqn (2) and (3):

[tex]T  = \frac{m5gR}{R} + mg = 6mg[/tex]

Thus tension at point Q is greater than the force at point P.

Answer: catching the ball is a better choice.

Explanation:

The collision of 2 objects involves involves large impact force since the force is inversely proportional to the time in which the momentum of the object changes.

Mathematically

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

If we catch the ball we increase the time in which the momentum of the ball is decreased thus the impact force that acts on us is lower as larger time is allowed for the ball to decrease it's momentum.

If we allow the ball to hit us the momentum of the ball changes in a short period of time thus applying a large impact force on our body thus increasing the chances of toppling.

Letting the ball bounce off of you is more likely to topple you.

This happens when two bodies hit each other with force. In this scenario, the ball has the force and we know that force is inversely proportional to the time in which the momentum of the object changes.

Letting the ball bounce off of you will decrease the time in which the momentum of the ball is increased thus the force that acts on us is higher which ius why it will most likely topple us.

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Final answer:

To calculate the atmospheric pressure, we can use the formula: Pressure = Density × Acceleration due to gravity × Height. Given the density of mercury, the height difference, and the acceleration due to gravity, we can calculate the atmospheric pressure.

Explanation:

The normal atmospheric pressure is 1.013 × 10^5 Pa. In this case, the height of a mercury barometer drops by 27.1 mm due to the approaching storm. To calculate the atmospheric pressure, we can use the formula:

Pressure = Density × Acceleration due to gravity × Height

Given that the density of mercury is 13.59 g/cm3, the height difference is 27.1 mm, and the acceleration due to gravity is approximately 9.8 m/s2, we can calculate the atmospheric pressure:

Pressure = (13.59 g/cm3)(9.8 m/s2)(27.1 mm)

Answer:

18 seconds

Explanation:

s = Displacement = 1.59×10³ m

a = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 0

v = Final velocity

t = Time taken

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1590=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1590\times 2}{9.81}}\\\Rightarrow t=18\ s[/tex]

Time taken by the raindrop to reach the ground neglecting air resistance is 18 seconds

Answer:

36 rev

Explanation:

See it in the pic

Answer:

Explanation:

If t be the time of fall of drops from a height of 181 s ,

181 = 1/2 g t²

t = 6 s approx

There are  4 drops in vertical line , one touching the floor, one  at 181 m height on the verge of falling down , and two drops in mid air.

The time interval between dripping of two consecutive drops

= 6 / 3

2 s

time duration of fall of second drop = 2 x 2 = 4 s

Position of drop below nozzle

= 1/2 g x 4²

= 78.4 m

b )

Time duration of fall of third drop

= 2 s

Position of drop below nozzle

= 1/2 9.8 x 2²

=  19.6 m

Answer:

Time, t = 3.04 hours

Explanation:

Given that,

Average speed of the marathon, v = 8.6 mi/h

Distance covered by the marathon runner, d = 26.22 mi

We need to find the time taken by the marathon runner to run that distance. Time taken is given by :

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{26.22\ mi}{8.6\ mi/h}[/tex]

t = 3.04 hours

So, the time taken by the marathon runner is 3.04 hours. Hence, this is the required solution.

Answer:

200.048 kg

Explanation:

Total Black dirt loaded in the pickup (A) = 260.5 kg = 260.500 kg

Quantity of the dirt that blew away (B )= 60452 g = 60.452 kg

Remaining quantity of the black dirt is

[tex]= 260.500 - 60.452[/tex]

[tex]= 200.048[/tex]

Thus, the amount of black dirt the man has when he reaches home = 200.048 kg.

Answer : The value of momentum is [tex]3.735\times 10^7kg.m/s[/tex] and m = 3.735 and n = 7

Explanation : Given,

Mass of high speed train = [tex]4.5\times 10^5kg[/tex]

Velocity of train = [tex]8.3\times 10^1m/s[/tex]

Formula used :

[tex]p=mv[/tex]

where,

p = momentum

m = mass

v = velocity

Now put all the given values in the above formula, we get:

[tex]p=(4.5\times 10^5kg)\times (8.3\times 10^1m/s)[/tex]

[tex]p=3.735\times 10^7kg.m/s[/tex]

Therefore, the value of momentum is [tex]3.735\times 10^7kg.m/s[/tex] and m = 3.735 and n = 7

Answer:

Part a)

[tex]L_o = 6.3181 cm[/tex]

Part b)

[tex]T = 131.3 ^oC[/tex]

Explanation:

Let the length of the rod at 0 degree Celsius is given as Lo

now we have

[tex]L = L_o( 1 + \alpha \Delta T)[/tex]

now we know that

[tex]L_o[/tex] = Length of rod at zero degree C

Part a)

[tex]6.3243 = L_o(1 + \alpha (16 - 0))[/tex]

[tex]6.3568 = L_o(1 + \alpha (100 - 0))[/tex]

now we have

[tex]\frac{6.3568}{6.3243} = \frac{1 + 100 \alpha}{1 + 16 \alpha}[/tex]

[tex]1.005(1 + 16 \alpha) = 1 + 100 \alpha[/tex]

[tex]83.918\alpha = 5.138\times 10^{-3}[/tex]

[tex]\alpha = 6.12 \times 10^{-5}[/tex]

now we have

[tex]L_o = 6.3181 cm[/tex]

Part b)

length of the rod is 6.3689 cm

now we have

[tex]L = L_o(1 + \alpha\Delta T)[/tex]

[tex]6.3689 = 6.3181(1 + \alpha \Delta T)[/tex]

[tex]6.3689 = 6.3181(1 + (6.12 \times 10^{-5})(T - 0))[/tex]

[tex]1.008 = 1 + (6.12 \times 10^{-5})T[/tex]

[tex]T = 131.3 ^oC[/tex]

Answer:

440 N

Explanation:

Force applied, F = 440 N

mass of crate, m = 170 kg

μs = 0.5, μk = 0.2

g = 9.81 m/s^2

The normal reaction acting on the crate, N = m g = 170 x 9.81 = 1667.7 N

The maximum value of static friction force acting on the crate

[tex]f_{s}=\mu _{s}N=0.5 \times 1667.7 = 833.85 N[/tex]

The maximum value of static friction force is more than the applied force so the crate does not move and teh applied force becomes friction force.

thus, the friction force acting on the crate is 440 N.

The frictional force exerted by the carpet on the crate is equal to the force of static friction.

To find the frictional force exerted by the carpet on the crate, we need to calculate the force of static friction first. The maximum force of static friction is given by fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the crate, which is given by N = mg. In this case, the normal force N = (170 kg)(9.81 m/s^2) = 1666.7 N.

Therefore, the maximum force of static friction is fs(max) = (0.5)(1666.7 N) = 833.35 N. Since the worker is pushing with a force of 440 N, which is less than the maximum force of static friction, the crate remains at rest and the frictional force exerted by the carpet on the crate is equal to the force of static friction, which is 833.35 N.

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Answer:

77 x 10⁻⁶ J

Explanation:

Work done by electric field

= charge x potential difference

= 7.7 x 10⁻⁶ x ( 15 - 5)

= 77 x 10⁻⁶ J

work done will be positive because direction of force and displacement are same.

Answer:

Part a)

the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge

Part b)

If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position

so this is Not stable equilibrium

Explanation:

Two charges are placed here on straight line are at 10 cm apart

here the two charges given are of opposite sign and hence the force on the third charge placed here on the same line will be zero where electric field is zero

Here electric field will be zero at the position near to the charge which is of small magnitude

so we will have

[tex]\frac{kq_1}{r^2} = \frac{kq_2}{(10 + r)^2}[/tex]

now we have

[tex]\frac{(1 \times 10^{-6})}{r^2} = \frac{(3\times 10^{-6})}{(10+ r)^2}[/tex]

[tex]10 + r = \sqrt3 r[/tex]

[tex]r = \frac{10}{\sqrt3 - 1} = 13.66 cm[/tex]

so the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge

Part b)

If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position

so this is Not stable equilibrium

Answer:

The position of near point is 1.68 m.

Explanation:

Given that,

Power = + 4.4 D

Object distance = 20 cm

We need to calculate the focal length

Using formula of power

[tex]P =\dfrac{1}{f}[/tex]

[tex]f=\dfrac{1}{P}[/tex]

[tex]f=\dfrac{1}{4.4}[/tex]

[tex]f=0.227\ m[/tex]

We need to calculate the image distance

Using formula of lens

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

[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}[/tex]

Put the value into the formula

[tex]\dfrac{1}{v}=\dfrac{1}{0.227}-\dfrac{1}{0.20}[/tex]

[tex]\dfrac{1}{v}=-\dfrac{135}{227}[/tex]

[tex]v=-1.68\ m[/tex]

Hence, The position of near point is 1.68 m.

Mary's near point without glasses is 22.73 cm, calculated by using the lens equation and the given lens power of her glasses with a knowledge of optics.

The question involves determining the near point location when the student, Mary, is not using her glasses. Mary has glasses with +4.4D converging lenses, and with the glasses, her near point is at 20 cm. To find the near point without glasses, we must use the lens equation:

1/f = 1/di + 1/do

Where f is the focal length of the lens, di is the image distance, and do is the object distance. For her glasses:  

The lens power P is +4.4D.

The focal length f is the inverse of power, f = 1/P.

Hence, f = 1/4.4 = 0.2273 meters (or 22.73 cm).

Since the near point with glasses is the closest distance at which she can see clearly, that means the image distance di (for her near point with glasses) is at infinity because the lenses correct her vision to normal. So:

1/f = 0 + 1/do

1/22.73 cm = 1/do

Therefore, the near point without glasses, do, is also 22.73 cm, since for the near point with glasses di is infinite - hence the person can see objects clearly at the near point with the glasses when the image is formed at infinity without the glasses.

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

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

Put  the value into the formula

[tex]-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2[/tex]

[tex]u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}[/tex]

[tex]u=-2.496\ m/s[/tex]

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

[tex]v = u+gt[/tex]

Put the value in the equation

[tex]-2.496=0-9.8\times t[/tex]

[tex]t =\dfrac{2.496}{9.8}[/tex]

[tex]t=0.254\ sec[/tex]

The total time is

[tex]t'=t+1.95[/tex]

[tex]t'=0.254+1.95[/tex]

[tex]t'=2.204\ sec[/tex]

We need to calculate the distance

Using equation of motion

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

Put the value into the formula

[tex]s=0+\dfrac{1}{2}\times9.8\times(2.204)^2[/tex]

[tex]s=23.80\ m[/tex]

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Answer:

a) 0.067 seconds

b) 112.5 m/s²

Explanation:

t = Time taken

u = Initial velocity = 7.5 m/s

v = Final velocity = 0

s = Displacement = 0.25 m

a = Acceleration

Eqaution of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-7.5^2}{2\times 0.25}\\\Rightarrow a=-112.5\ m/s^2[/tex]

b) The deceleration of the football player is 112.5 m/s²

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-7.5}{-112.5}\\\Rightarrow t=0.067\ s[/tex]

a) The collision lasts for 0.067 seconds