At what speed does a 2000 kg compact car have the same kinetic energy as a 18000 kg truck going 21 km/hr?

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Tension of the line = the weight + force from pulling up the fish

30 N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6 a

a = 1.7333333 m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

Answer:

C. a liter of salt water.

Explanation:

Defination of Solution =>

a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).

Answer:

Please check the explanation

Explanation:

The best two methods will be:

Factor by grouping

Factor by grouping deals with establishing a smaller groups from each term.

[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)[/tex]

[tex]8\:=\:\:2\cdot 2\cdot 2[/tex]

Therefore, the expression becomes

[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)-\left(2\cdot \:2\cdot \:2\right)[/tex]

Now factor out the greatest common factor (GCF) which is 2

         [tex]=\:2\left(3\cdot \:\:3x^2-\left(2\right)\left(2\right)\right)[/tex]

           [tex]=2\left(9x^2-2\cdot \:2\right)[/tex]

            [tex]=2\left(9x^2-4\right)[/tex]

Factor out the GCF

Given the expression

[tex]18x^2-8\:\:\:[/tex]

factor out common term 2

[tex]=2\left(9x^2-4\right)[/tex]

[tex]=2\left(3x+2\right)\left(3x-2\right)[/tex]          ∵ [tex]Factors\:\:\left(9x^2-4\right)=\left(3x+2\right)\left(3x-2\right)[/tex]

Answer:should be a matter of vector analysis.

Pulling above the horizontal has less surface area for the opposing friction

Explanation:

Answer:

This question is incomplete.

Explanation:

This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below

V = IR

where V is the voltage. I is the current and R is the resistance

R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......

Answer:

[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]

Explanation:

[tex]F_1=0.072\ \text{N}[/tex]

[tex]F_2=0.115\ \text{N}[/tex]

r = Distance between shells = 40.4 cm

[tex]q_1[/tex] and [tex]q_2[/tex] are the charges

[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]

Force is given by

[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]

[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]

[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]

Substituting the above value of [tex]q_1[/tex] we get

[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]

[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]

[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]

Since we know [tex]q_1<q_2[/tex]

[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].

Answer:

0.25m

Explanation:

Given parameters:

Spring constant , K  = 120N/m

Work done  = 3.75J

Unknown:

magnitude of extension = ?

Solution:

To solve this problem;

           Work done  = [tex]\frac{1}{2}[/tex]kx²  

K is the spring constant

x is the extension

               3.75  =  [tex]\frac{1}{2}[/tex] x 120x²

               3.75  = 60x²

                x²  = 0.06

                x = √0.06  = 0.25m

Answer:A

Explanation:

Answer:

A. -2.83 x 10-7N

Explanation:

Answer:

x = 1127 [m]

Explanation:

In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.

[tex]v_{f} =v_{o} -a*t[/tex]

where:

Vf = final velocity = 9 [m/s]

Vo = initial velocity = 55 [m/s]

a = acceleration o desacceleration [m/s²]

t = time = 49 [s]

Now replacing:

9 = 55 - a*49

a*49 = 55 + 9

a = 1.306 [m/s²]

Note: The negative sign in the above equation means that the speed decreases.

Now using the second equation.

[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]

(9)² = (55)² - 2*(1.306)*x

2944 = 2.612*x

x = 1127 [m]

Answer:

349 Hz

Explanation:

We are told that the guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beat/s when sounded with a 357 Hz tuning fork.

This means that for the 352 Hz tuning fork, the vibrational frequency is;

f = 352 ± 3

f = (352 + 3) or (352 - 3)

f = 355 Hz or 349Hz

For the 357 Hz tuning fork, the vibrational frequency is;

f = 357 ± 8

f = (357 + 8) or (357 - 8)

f = 365 Hz or 349 Hz

In both cases, 349 Hz is common;

Thus, the vibrational frequency of the string = 349 Hz

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

Answer:

a

[tex]\lambda_{long} = 288.5 \ nm[/tex]

b

The velocity is  [tex]v = 3.7 *0^{5} \ m/s[/tex]

Explanation:

From the question we are told that

   The work function of Zinc is  [tex]W = 4.3 eV[/tex]

Generally the work function can be mathematically represented as

     [tex]E_o = \frac{hc}{\lambda_{long}}[/tex]

=>   [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]

Here  h is the Planck constant with the value  [tex]h = 4.1357 * 10^{-15} eV s[/tex]

  and c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

     [tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]

=>  [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]

=>  [tex]\lambda_{long} = 288.5 \ nm[/tex]

Generally the kinetic energy of the emitted electron is mathematically represented as

      [tex]K = E -E_o[/tex]

Here  E is the energy of the photon that strikes the surface

So

    [tex]E- E_o = \frac{1}{2} m * v^2[/tex]

Here m is the mass of electron with value  [tex]m = 9.11*10^{-31 } \ kg[/tex]

Generally  [tex]1 ev = 1.60 *10^{-19} \ J[/tex]

=>   [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]

=>    [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]

=>    [tex]v = 3.7 *0^{5} \ m/s[/tex]

Answer:

both

Explanation:

baby for fun, toddler for vengence

Answer:

delivery truck

Explanation:

because i got it right

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

[tex]F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}[/tex]

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

[tex]I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A[/tex]

Therefore, the bottom current is 12.8 A to the right.

Answer:

The amplitude of the motion is 0.0286 m.

Explanation:

Given;

mass of the object, m = 0.2 kg

spring constant, k = 120 N/m

maximum speed of the simple harmonic motion, [tex]V_m[/tex] = 0.70 m/s

The amplitude A of the motion is given by;

[tex]V_m = \omega A\\\\[/tex]

where;

ω is the angular velocity given as;

[tex]\omega = \sqrt{\frac{k}{m} }\\\\\omega = \sqrt{\frac{120}{0.2} }\\\\\omega =24.5 \ rad/s[/tex]

Now, substitute the value of angular velocity and solve the amplitude;

[tex]V_m = \omega A\\\\A = \frac{V_m}{\omega}\\\\A = \frac{0.7}{24.5}\\\\A = 0.0286 \ m[/tex]

Therefore, the amplitude of the motion is 0.0286 m.

Answer:  

Correct answer to this is: A

Hopefully, this helped out a bit :)  

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

[tex]\tau = I\alpha[/tex]

Where [tex]\tau[/tex] is the torque

[tex]I[/tex] is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

But, the angular acceleration is given by

[tex]\alpha = \frac{\omega}{t}[/tex]

Where [tex]\omega[/tex] is the angular speed

and [tex]t[/tex] is time

Then, we can write that

[tex]\tau = \frac{I\omega}{t}[/tex]

Hence,

[tex]\omega = \frac{\tau t}{I}[/tex]

Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].

Here, The torque is given by,

[tex]\tau = rF[/tex]

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ [tex]\tau = 3.00 \times 195[/tex]

[tex]\tau = 585[/tex] Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

[tex]I = \frac{1}{2}MR^{2}[/tex]

Where M is the mass and

R is the radius

∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]

[tex]I = 1462.5[/tex] kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

[tex]\omega = \frac{\tau t}{I}[/tex]

[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]

[tex]\omega = 0.82[/tex] rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

Answer:

its 0.5 for all i beleive

Explanation:

Answer:

The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²

Explanation:

Given;

distance traveled in the given time = 200 m

time to cover the distance, t = 29.6 s

speed of the runner, v = d / t

v = 200 / 29.6

v = 6.757 m/s

The centripetal acceleration of the runner is given by;

[tex]a_c = \frac{V^2}{r}[/tex]

where;

r is the radius of the circular arc, given as 50 m

Substitute the givens;

[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]

Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².

Answer:

Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.

Explanation:

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.

Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.

Droughts can last months or years, although they can be proclaimed in as little as 15 days.

It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.

Hence Precipitation and droughts are the specific changes in two climate variables.

To learn more about the drought refer to the link;

https://brainly.com/question/26693108

Answer:

D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]

Explanation:

The statement is not correctly written, the correct form is now described:

A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?

From classic theory on Magnetism, we remember that the magnetic force  exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:

[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)

Where:

[tex]q[/tex] - Electric charge, measured in coulombs.

[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.

[tex]\vec B[/tex] - Magnetic field, measured in teslas.

If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:

[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]

[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]

[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]

Which corresponds to option D.

Answer:

it would be 49.03325 Newton.

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

[tex]F_D = \frac{1}{2}C\rho Av^2[/tex]

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

[tex]F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N[/tex]

The downward force of the ball due to its weight and that of the ant is given by;

[tex]F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N[/tex]

The net downward force experienced by the ball is given by;

[tex]F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N[/tex]

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

Answer:

I'm pretty sure its A

Explanation:

because its a reflection- Hope you get a good grade!

Answer:

Explanation:

Step one:

given data

mass = 7kg

acceleration =2m/s^2

time= 9seconds

acceleration = velocity/time

velocity= acceleration *time

velocity=2*9

velocity= 18m/s

distance moved= velocity* time

distance= 18*9

distance=162m

we also know that the force on impulse is given as

Ft=mv

F=mv/t

F=7*18/9

F=126/9

F=14N

work done = Force* distance

work done=14*162

work=2268Joules

work= 2.27kJ

Answer:

i think it should be to make life easier