A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.

Answer: The correct statement is:

--> The voltage across each resistor is the same.

Explanation:

RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:

--> Resistors in parallel and

--> Resistors in series

RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

Solution :

Part A .

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km

           [tex]y[/tex] component is = 4 x cos (29°) = 3.498 km

Part B

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]

So the [tex]x[/tex] component is = -2 cm/s

           [tex]y[/tex] component is = 0

Part C

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]

           [tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]

The x- and y-components of the vectors  is mathematically given as as follows for each Part respectively

x= -1.939 km, y= 3.498 km

x= -2 cm/s, 0

y=, x= -7.6412m/s^2, -10.517m/s^2

What are the x- and y-components of the vectors?

Question Parameters:

Generally, we follow a basic principle where

x component= Fsin\theta

y component= Fcos\theta

Therefore

For A

x component is

x= -4 x sin (29°)

x= -1.939 km

 y component is

y= 4 x cos (29°)

y= 3.498 km

For B

x component is

x= -2 cm/s            

y component is

y= 0

For C

x component is

x= -13 x sin (36°)

x= -7.6412m/s^2      

y component is

y= -13 x cos (36°)

y= -10.517m/s^2  

Read more about Cartession co ordinate

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Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex]  Filling in:

[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

[tex]E_0=1.5033×10^{-10}\:\text{J}[/tex]

Explanation:

The rest energy [tex]E_0[/tex] of a proton of mass [tex]m_p[/tex] is given by

[tex]E_0 = m_pc^2[/tex]

[tex]\:\:\:\:\:\:\:=(1.6726×10^{-27}\:\text{kg})(2.9979×10^8\:\text{m/s})^2[/tex]

[tex]\:\:\:\:\:\:\:=1.5033×10^{-10}\:\text{J}[/tex]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

Answer:

b.have frequencies that are integer multiples of the frequency of the complex waveform

Explanation:

Please correct me if I am wrong

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

Answer:

False

Explanation:

You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.

Answer:

The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ [tex]P=qd[/tex]

By putting the values, we get

⇒     [tex]=0.50\times 900[/tex]

⇒     [tex]=(0.50\times 10^{-9})\times 0.9[/tex]

⇒     [tex]=4.5\times 10^{-10} \ Cm[/tex]  

Answer:

The dipole moment is 4.5 x 10^-10 Cm.

Explanation:

Charge on each ball, q = 0.5 nC

Length, L = 900 mm = 0.9 m

The dipole moment is defined as the product of either charge and the distance between them.

It is a vector quantity and the direction is from negative charge to the positive charge.

The dipole moment is

[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

Answer:

c)  N_s / N_p = 115.15

Explanation:

Let's look for the voltage in the secondary, they do not indicate the power dissipated

          P = V_s i

          V_s = P / i

          V_s = 76 / 5.5 10⁻³

          V_s = 13.818 10³ V

the relationship between the primary and secondary of a transformer is

           [tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]

           [tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]

           Ns / Np = 13,818 10³ /120

           N_s / N_p = 115.15

Answer:

Test 1

1.True

2.True

3.True

4.False

5.True

6.True

7.False

8.True

9.True

10.True

yung iba nasa pic

Answer:

-45,597.07

Explanation:

if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh

Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

Answer:

(i)8m/s²(ii)40m/s

Explanation:

according to the formula

½at²=s.

then substituting the data

½a•5²=100

a=8m/s²

v=at=8•5=40m/s

Answer:

(I)

[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

         [tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

            [tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]

             1 / p = 0.107375

             p = 93.13 cm

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

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

By rearrangement and making (v) the subject of the above formula:

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

replacing the given values:

[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]

[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]

v = 0.857 cm

Answer:

have an increased resistance

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

[tex]M = \frac{q}{p}[/tex]

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]

Now using thin lens formula:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]

f = 1 m

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]

So, the approximate pressure is equal to 40.69 atm.

Answer:

Explanation:

Answer:

3.464 seconds.

Explanation:

We know that we can write the period (the time for a complete swing) of a pendulum as:

[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]

Where:

[tex]\pi = 3.14[/tex]

L is the length of the pendulum

g is the gravitational acceleration:

g = 9.8m/s^2

We know that the original period is of 2.00 s, then:

T = 2.00s

We can solve that for L, the original length:

[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]

So if we triple the length of the pendulum, we will have:

L' = 3*0.994m = 2.982m

The new period will be:

[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]

The new period will be 3.464 seconds.

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field