A cube with sides of area 48 cm^2 contains a 28.7 nanoCoulomb charge. Find the flux of the electric field through the surface of the cube in unis of Nm^2/C.
Please conceptually explain this question answer to me! Thanks!!

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:

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:

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:

(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.

https://brainly.com/question/26472013

#SPJ11

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:

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:

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:

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.

The magnitudes of vectors A and B can be determined by resolving each vector into its x and y components, setting up equations for their sum, and solving the equations simultaneously.

To find the magnitudes of vectors A and B when their sum with vector C results in a zero vector (A + B + C = 0), we can break down each vector into its x and y components and then solve the component equations separately.

Since vector C points in the negative y-direction and has a magnitude of 21.0 m, its components are Cx = 0 m and Cy = -21.0 m.

Vector A has a positive y-component and makes an angle of α = 43.4° with the positive x-axis. Thus, the components of A are Ax = A cos(α) and Ay = A sin(α).

Vector B also has a positive y-component and makes an angle of β = 27.7° with the negative x-axis, which is the same as 180° - β with the positive x-axis. So, B's components are Bx = -B cos(180° - β) = -B cos(β) and By = B sin(180° - β) = B sin(β).

The equations for the x and y components of the vector sum being zero are:

Substituting the component expressions in, we get:

These are two equations with two unknowns (the magnitudes A and B), which can be solved simultaneously to find the values of A and B. The solution involves elementary algebra and the use of trigonometric identities.

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:

The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].

Explanation:

Given that,

Speed [tex]v= 5.05\times10^{5}\ m/s[/tex]

Time [tex]t= 3.90\times10^{-7}\ s[/tex]

Angle = 45°

We need to calculate the acceleration

Using equation of motion

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

[tex]5.05\times10^{5}=0+a\times3.90\times10^{-7}[/tex]

[tex]a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}[/tex]

[tex]a=1.29\times10^{12}\ m/s^2[/tex]

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

[tex]F= ma=qE[/tex]

[tex]ma = qE[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

Put the value into the formula

[tex]E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}[/tex]

[tex]E=1.35\times10^{4}\ N/C[/tex]

Hence, The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].

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.

https://brainly.com/question/31444153

#SPJ3

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:

[tex]OF=\rho _wVg[/tex]

Explanation:

The bouyancy force an object feels (OF) when submerged ina fluid is always the weight of the liquid the object displaces. This weight will be the mass of fluid displaced ([tex]m_d[/tex]) multiplied by the acceleration of gravity g. The mass displaced will be the density of the fluid [tex]\rho_w[/tex] multiplied by the volume of fluid displaced [tex]V_d[/tex]. If the object is fully submerged, then this volume will be the same as the volume of the object V. We write all these steps in equations:

[tex]OF=m_dg=\rho _wV_dg=\rho _wVg[/tex]

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:[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]

Explanation:

Given

v=initial velocity

resisting acceleration =cv

also gravity is opposing the upward motion

Therefore distance traveled during upward motion

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

Where a=cv+g

[tex]0-v^2=2(cv+g)s[/tex]

[tex]s=\frac{v^2}{2(cv+g)}[/tex]

Now let v_0 be the velocity at the ground

[tex]v^2_0-0=2(g-vc)s[/tex]

substituting s value

[tex]v^2_0=v^2\frac{g-vc}{(cv+g)}[/tex]

[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]

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:

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:

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: 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.

Read more about Collision here https://brainly.com/question/24915434

Answer:48.52 kJ

Explanation:

Given

Resistance[tex]=100 \Omega [/tex]

temperature increases from [tex]10^{\circ}C to 20^{\circ}C[/tex]

Voltage=50 V

Heat given(H)[tex]=\frac{V^2t}{R}[/tex]

Where V=voltage applied

t=time

R=Resistance

[tex]H=\frac{50^2\times 60\times 60}{100}=90 kJ[/tex]

Heat absorbed by water is

[tex]Q=mc(\Delta T)[/tex]

where

m=mass of water

c=specific heat of water

[tex]\Delta T[/tex]=change in temperature

[tex]Q=1000\times 4.184\times (20-10)=41.48 kJ[/tex]

Therefore  90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.

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:

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:

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:

The force required by the brakes is [tex]3.67\times 10^{3} N[/tex]

Solution:

As per the question:

Mass of the truck, M = 2200 kg

Height, h = 50 m

distance moved by the truck before stopping, x = 300 m

[tex]g = 10 m/s^{2}[/tex]

Force required by the brakes to stop the truck, [tex]F_{b}[/tex] can be calculated by using the law of conservation of energy.

Now,

Kinetic Energy(K.E) downhil, K.E = reduction in potential energy, [tex]\Delta PE[/tex]

[tex]\Delta PE = Mgh = 2200\times 10\times 50 = 1100 kJ[/tex]

Work done is provided by the decrease in K.E,i.e.,  change in potential energy.

W = [tex]F\times x = 300 F = 1100\times 10^{3}[/tex]

F = [tex]3.67\times 10^{3} N[/tex]

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:

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:

22.93 seconds

Explanation:

s = Displacement = 3000 - 420 = 2580 m = The distance she will free fall

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 2580=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2580\times 2}{9.81}}\\\Rightarrow t=22.93\ s[/tex]

Time taken by the skydiver to cover the distance between 3000 m to an altitude of 420 m neglecting air resistance is 22.93 seconds

Answer:

A  = 3.55 m

Explanation:

given,

depth of water = 20 m                

Amplitude of wave at 20 m depth = 1.1 m        

Amplitude at the depth  of 6.2 m              

amplitude is inversely proportional to depth of water

so,                                      

[tex]\dfrac{A_1}{A_2} = \dfrac{d_2}{d_1}[/tex]

[tex]A_2= A_1\dfrac{d_1}{d_2}[/tex]

[tex]A_2= 1.1\dfrac{20}{6.2}[/tex]

A  = 3.55 m

hence, the amplitude of the wave at the depth of 6.2 m is 3.55 m.

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]