PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Explanation:

if the current is 1A

V=iR

V= 1 × 8

V = 8volts

Answer:

12558 J

Explanation:

Please do mark as brainliest. Hope this helps! :)

Answer:

you need 7 points and the other team just needs to stop you from scoring

Explanation:

Answer:

(a) a = 2.44 m/s²

(b) s = 63.24 m

Explanation:

(a)

We will use the second equation of motion here:

[tex]s = v_it+\frac{1}{2}at^2[/tex]

where,

s = distance covered = 47 m

vi = initial speed = 0 m/s

t = time taken = 6.2 s

a = acceleration = ?

Therefore,

[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]

a = 2.44 m/s²

(b)

Now, we will again use the second equation of motion for the complete length of the inclined plane:

[tex]s = v_it+\frac{1}{2}at^2[/tex]

where,

s = distance covered = ?

vi = initial speed = 0 m/s

t = time taken = 7.2 s

a = acceleration = 2.44 m/s²

Therefore,

[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]

s = 63.24 m

Answer:

When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work.

Explanation:

Hope this helps you

Answer:

V₁ = V = 120 V

Explanation:

Such a combination of capacitors in which;

1- Potential difference across each capacitor is the same

2- Total charge is distributed amongst the capacitors

; is called Parallel Combination.

Therefore, in this case, the potential difference across each capacitor will also be the same. Because the capacitors are connected in parallel here. So the voltage across 3 μF capacitor will be the same as the voltage across the 6 μF capacitor and they both will be equal to the total potential difference.

V₁ = V = 120 V

Answer:

hope u can understand the method

Answer:

the answer of this question is true

Solution :

Mass is varied keeping frequency constant.

Wavelength, λ  [tex]$=\frac{2l}{n}$[/tex]

where length of spring = l

           number of segments = n

Velocity, v = λ x f

                 = [tex]$\sqrt{\frac{T}{\mu}}$[/tex]

[tex]$\mu $[/tex] =  mass density, T = tension in string

[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]

[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex]  , n = 2

[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]

[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]

[tex]$\Rightarrow m = 2.619\ kg$[/tex]

Therefore, μ = 0.002785 kg/ m

Frequency is varied keeping T constant

[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]

[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]

[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]

f = 210 Hz

Answer:

3.26mph

Explanation:

To calculate speed, use the formula distance/time. In this case, just divide 14.7  by 4.5.

The speed with which the cat runs towards its food bowl is 7.3 Miles per hour.

Given the data in the question

Speed of the cat; [tex]v = \ ?[/tex]

Speed is the time rate at which an object is moving along a path

Speed is the time rate at which an object is displaced.

Speed = Distance / Time

[tex]v = s / t[/tex]

We substitute our values into the equation

[tex]v = \frac{14.7m}{4.5s}\\\\v = 3.267 m/s\\\\v = 7.3mph[/tex]

Therefore, the speed with which the cat runs towards its food bowl is 7.3 Miles per hour.

Learn more: https://brainly.com/question/680492

Answer:

a

because the mechanical wave is when it goes over and over again

Answer:

The answer is a like i said 3hrs ago i dont know if this guy copied me tbh

Explanation:

Answer:

m=29.6kg

Explanation:

your welcome:) have a great day

Answer:

R = v^2 sin 2 theta / g

The range provides the distance a projectile can travel

R(max) = v^2 / g    if theta = 45 deg

R = 2.7^2 / 9.8 = .74 m

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

[tex]2gh = v_f^2 - v_i^2\\[/tex]

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

[tex](2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s[/tex]

Now, we will apply the law of conservation of momentum:

[tex]m_1v_1 = m_2v_2[/tex]

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

[tex](3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s[/tex]

Now, we again use the third equation of motion for the upward motion of the ball:

[tex]2gh = v_f^2 - v_i^2\\[/tex]

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

[tex](2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\[/tex]

h = 3.5 m

Answer: 4.94cm³

Explanation:

Data;

ρ = 1.59g/cm³

mass = 7.85g

volume = ?

density = mass / volume

ρ = m / v

v = m / ρ

v = 7.85 / 1.59

v = 4.94cm³

Answer:

0.063 Kg

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 10 Hz

Spring constant (K) = 250 N/m

Mass (m) =?

Next, we shall determine the period of oscillation. This can be obtained as follow:

Frequency (f) = 10 Hz

Period (T) =?

T = 1/f

T = 1/10

T = 0.1 s

Finally, we shall determine the mass of the spring. This can be obtained as follow:

Spring constant (K) = 250 N/m

Period (T) = 0.1 s

Pi (π) = 3.14

Mass (m) =?

T = 2π√(m/K)

0.1 = 2 × 3.14 × √(m/250)

0.1 = 6.28 × √(m/250)

Divide both side by 6.28

0.1 / 6.28 = √(m/250)

Take the square of both side.

(0.1 / 6.28)² = m/250

Cross multiply

m = (0.1 / 6.28)² × 250

m = 0.063 Kg

Therefore, the mass of the spring is 0.063 Kg.

Answer: Different types of telescopes usually don't take simultaneous readings. Space is a dynamic system, so an image taken at one time is not necessarily the precise equivalent of an image of the same phenomena taken at a later time. And often, there is barely enough time for one kind of telescope to observe extremely short-lived phenomena like gamma-ray bursts. By the time other telescopes point to the object, it has grown too faint to be detected.

Explanation: Trust me

Answer:

Explanation:

Whats the entire question?

Answer:

C, Red has the longest one

c red

red has longest wavelength

amnh dot org

This is true I think

It applies to Newton's Laws

it's true because it's a part of newtons law

Answer:

J = 1600 kg-m/s

Explanation:

Given that,

The mass of charging bull rams, m = 800 kg

Initial speed, u = 5 m/s

Final speed, v = 3 m/s

We need to find the impulse the bull  experience by smashing the fence. Let it is J. We know that, impulse is equal to the change in momentum such that,

J = m(v-u)

Put all the values,

J = 800(3-5)

= 800(-2)

= -1600 kg-m/s

Hence, the magnitude of impulse is equal to 1600 kg-m/s.

Answer:

(a) Energy Density = 160.94 J/m³

(b) Energy Stored = 0.192 J

Explanation:

(a)

The energy density of the magnetic field inside the solenoid is given by the following formula:

[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]

where,

B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]

Therefore,

[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]

where,

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 1300

I = current = 8.15 A

L = length = 66.2 cm = 0.662 m

Therefore,

[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]

Energy Density = 160.94 J/m³

(b)

Energy Stored = (Energy Density)(Volume)

Energy Stored = (Energy Density)(Area)(L)

Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)

Energy Stored = 0.192 J

Answer:

Hello! Your answer would be, A) True

Explanation:

Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

        Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

        Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

       Em₀ = Em_f

       ½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

       ½ k x² - m₁ g y = ½ m₂ v²

       v² = [tex]\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y[/tex]

let's calculate

        v² = [tex]\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)[/tex]

        v² = 2.7 + 5.23

        v = √7.927

        v = 2,815 m / s

using of significant figures

        v = 2.82 m/s

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency [tex]\zeta[/tex] = 3%

where;

[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]

[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]

[tex]Q_{supplied } = 66.66 \ MW[/tex]

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]

Also;

[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]

[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]

[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]

[tex]LMTD = \dfrac{8}{In (5)}[/tex]

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

[tex]Q_H = UA (LMTD)[/tex]

where;

U = overall heat coefficient given as 1200 W/m².K

[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]

The mass flow rate:

[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]

Answer:

the maximum allowable current is 7302.967  amperl

Explanation:

The computation of the maximum allowable current is shown below;

Force F = mean ÷ 4π 2 I_1 I_2 ÷d  × ΔL

200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5

200 = 3.75 × 10^-6 I^2

I = √200 ÷ √ 3.75 × 10^-6

= 7302.967  amperl

Hence, the maximum allowable current is 7302.967  amperl

Basically we applied the above formula

Solution :

Given :

M = 0.35 kg

[tex]$m=\frac{M}{2}=0.175 \ kg$[/tex]

Total mechanical energy = constant

or [tex]$K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$[/tex]

But [tex]$K.E._{top} = 0$[/tex] and [tex]$P.E._{bottom} = 0$[/tex]

Therefore, potential energy at the top = kinetic energy at the bottom

[tex]$\Rightarrow mgh = \frac{1}{2}mv^2$[/tex]

[tex]$\Rightarrow v = \sqrt{2gh}$[/tex]

      [tex]$=\sqrt{2 \times 9.8 \times 0.35}$[/tex]      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

[tex]$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$[/tex]

here, [tex]$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$[/tex]

∴ [tex]$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}[/tex]

      [tex]$=\frac{0.175}{0.525}+0$[/tex]

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s