Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s. Use the initial direction of the cars motion as the positive direction.
What impulse, in kilogram meters per second, is imparted to the car by this force?
Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.
Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

Answer:

Explanation:

The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point

Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima  is π radian .

Answer:

Velocity of bear (v) = 20 m/s

Explanation:

Given:

Mass of bear (m) = 500 kg

Mechanical kinetic energy (K.E) = 100,000 J

Find:

Velocity of bear (v) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2(m)(v)²

100,000 = 1/2(500)(v)²

200,000 = 500(v)²

400 = (v)²

Velocity of bear (v) = 20 m/s

Answer:

processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Explanation:

The temperature of planet Earth is due to two main types of process, internal and external.

Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect

External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.

These are the two terms that heat the Earth

In addition there are several processes so the planet loses energy,

* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission

* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system

* liquid (water) system that absorbs large amounts of heat to change state and temperature.

These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.

Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value

The amount of charge that passes per unit time is called electric current .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

Answer:

He was going at 54.5mph

Explanation:

Measuring distances in feet and time (starting with the brakes being applied) in seconds,

we have a(t) = −16, s(0) = 0, and v(T) = 0, s(T) = 200,

where T is the time of braking.

So Finding antiderivatives, v(t) = −16t + A and so s(t) = −8t²+ At +B. Since s(0) = 0,

we have B = 0. Since v(T) = 0,

we have T = A /16.

Putting this into s() gives

200 = −8(A/ 16)² +A· (A /16) = A²/32.

This gives A2 = 6,400, so A = 80 = v(0). That is, the initial speed was 80 ft/s ≈ 54.5 mph.

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

Answer:

a. 8.465 m/s

b.22.3659

Answer:

The value of the inductance is 0.955 mH

Explanation:

Given;

frequency of the oscillator, f = 18 kHz = 18,000 Hz

the peak current, I₀ = 70 mA = 0.07 A

the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V

The root mean square current is given as;

[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]

[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]

Inductive reactance is given by;

[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]

Inductance is given by;

[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]

L = 0.955 mH

Therefore, the value of the inductance is 0.955 mH

The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.

Given the following data:

To determine the value of the inductance (L):

First of all, we would find the root mean square (rms) current by using the formula:

[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]

Next, we would calculate the inductive reactance of the oscillator by using the formula:

[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]

Now, we can solve for the value of the inductance (L):

[tex]L = \frac{X_L}{2\pi f}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]

L = [tex]9.55 \times 10^{-4}[/tex] Henry.

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Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]

So, the ratio of masses X/Y is 0.0847

Answer:

34.58 N

Explanation:

From the question,

Tension in the cord =  Weight of the solid metal- Upthrust

T = W-U................... Equation 1

But,

W = mg

Where m = mass of solid, g = acceleration due to gravity.

Given: m = 4.7 kg, g = 9.8 m/s²

W = 4.7(9.8) = 46.1 N.

U = (Density of water×Volume of water displaced.)gravity

U = D×V×g.

But,

V = Mass of solid/density of solid

V = 4.7/4000

V = 1.175×10⁻³ m³.

Therefore,

U = 1.175×10⁻³(1000)(9.8)

U = 11.52 N

Therefore,

T = 46.1-11.52

T = 34.58 N

Answer:

The maximum emf induced in the loop is 0.132 Volts

Explanation:

Given;

radius of the circular loop, r = 9.5 cm

intensity of the wave, I = 0.0295 W/m²

wavelength, λ = 6.40 m

The intensity of the wave is given as;

[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]

where;

B₀ is the amplitude of the field

c is the speed of light = 3 x 10⁸ m/s

μ₀ is permeability of free space = 4π x 10⁻⁵ m/A

[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]

Area of the circular loop;

A = πr²

A = π(0.095)²

A = 0.0284 m²

Frequency of the wave;

f = c / λ

f = (3 x 10⁸) / (6.4)

f = 46875000 Hz

Angular velocity of the wave;

ω = 2πf

ω = 2π(46875000)

ω = 294562500 rad/s

The maximum induced emf is calculated as;

emf = B₀Aω

       = (1.572 x 10⁻⁸)(0.0284)(294562500)

       = 0.132 Volts

Therefore, the maximum emf induced in the loop is 0.132 Volts

Answer:

(a) 0.59 ms

(b) 0.15 ms

(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.

Explanation:

(a) distance between ears = 20 cm = 0.2 m

speed of sound generated = 340 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                        = [tex]\frac{0.2}{340}[/tex]

                        = 5.8824 × [tex]10^{-4}[/tex]

                        = 0.59 ms

The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.

(b) distance between ears = 20 cm = 0.2 m

speed of sound in water = 1530 m/s

time = ?

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]

                         = [tex]\frac{0.2}{1530}[/tex]

                         = 1.4815 × [tex]10^{-4}[/tex]

                         = 0.15 ms

The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.

(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.

A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds

B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds

C) The significance about the time interval of the two situations is that;

Transmission of sound varies with different mediums.

A) We are given;

Distance between the two ears; d = 20 cm = 0.2 m

Speed of sound; v = 340 m/s

Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;

t = d/v

t = 0.2/340

t = 0.588 × 10⁻³ seconds

B) We are now told that the speed of sound in water is 1530 m/s. Thus;

t = 0.2/1530

t = 0.131 × 10⁻⁵ seconds

C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.

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Answer:

sphere #1 carries positive charge and #2 carries negative charge

This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]

Where:

[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.

[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]

By using the definition of momentum, the expression is therefore expanded:

[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]

Where:

[tex]m[/tex] - Mass of the ball, measured in kilograms.

[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.

If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:

[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]

[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]

[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is:

[tex]\Delta p = p_{f}-p_{o}[/tex]

[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]

[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Explanation:

From the question we are told that

   The  width of the slit is  [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]

     The angle is  [tex]\theta = 11^o[/tex]

The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as

        [tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]

Where [tex]\beta[/tex] is mathematically represented as

        [tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]

substituting values

      [tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]

     [tex]\beta = 708.1 \ rad[/tex]

So

  [tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]

   [tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

Answer:

15.01 m/s

Explanation:

Applying newtons equation of motion,

v² = u²+2gs.................. Equation 1

Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.

Given: u = 21 m/s, s = 11 m

Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).

Substitute these values into equation 1

v² = 21²-2(11)(9.8)

v² = 441-215.6

v² = 225.4

v = √225.4

v = 15.01 m/s

Answer:

B. only when the current in the first coil changes

Explanation:

This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced

Answer:

Explanation:

Using equations of motion:

(a)

v=u+at

∴0=19.4−8.57t

∴t=19.4/8.57

=2.3s

B. Using s= ut + 1/2 at²

19.4(2.3)-1/28.57(2.3)²

= 21.92rad

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

Answer:

3.26m

Explanation:

See attached file

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

Answer:

Potential gradient = 1.6 v/m  

E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l

l=1.2\1.6

=3\4

3\4=0.75m

convert m into cm 0.75x10

0.75m=75 cm

ans=75 cm

Explanation:

Answer:

1) 14 kW

2) 4.67 x 10^-5 N

Explanation:

Area of solar panel = 10 m^2

Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

P = 1.4 kW/m^2  x  10 m^2 = 14 kW

b) For a perfect absorbing surface, the radiation pressure is given as

p = I/c

where p is the radiation pressure

I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N

Answer:

[tex]I=2.67\times 10^{-6}\ W/m[/tex]

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]

So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].

Answer:

   ΔP = - 0.262 ρ v₁²

Explanation:

This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

suppose the pipe is horizontal, therefore

           y₁ = y₂

         P₁ -P₂ = ½ ρ (v₁² -v₂²)             1

Now let's use the continuity equation

         v₁ A₁ = v₂ A₂                              2

The area of ​​a circle is

         A = π r²

also the radius is half the diameter r = d /2

         A₁ = π d₁² / 4

let's substitute in equation 2

         v1 π d₁² / 4 = v2 π d₂² / 4

          v₁ d₁² = v₂ d₂²

          v₂ = v₁ d₁² / d₂²

let's substitute in equation 1

          P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )

Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1

           d₂ = D -0.1D = 0.9 D

we substitute in the previous equation

          P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)

          P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )

          P1 - P2 = ½ ρ v12 (- 0.524)

          ΔP = - 0.262 ρ v₁²

in this solution we assume that the data in zone I is known

Answer:

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V