A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.

Answer: no u

Explanation: no u

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

Answer:

82.2 mL

Explanation:

The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;

C1V1=C2V2

Where;

C1= concentration of stock solution

V1= volume of stock solution

C2= concentration of dilute solution

V2= volume of dilute solution

V2= C1V1/C2

V2= 1.48 × 55.6/ 1.0

V2= 82.2 mL

Answer:

a. Moon around Earth.

Explanation:

Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.

The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.

The object that should exhibit the longest orbital period is option a. Moon around Earth.

Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.

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Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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(a) The car is undergoing an acceleration of

[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]

so that in 9.02 s, it will have covered a distance of

[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]

(b) The wheels have average angular velocity

[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]

where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

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Answer:

it is either constant or zero

Explanation:

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

Answer:

0.09 x10^-10m

Explanation:

Using wavelength=( 12.27 A)/√V

= 12.27 x 10^-10/ √1.6x10^2

= 0.09x10^-10m

Answer:

covalent

Explanation:

covalent bonds share electrons

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

Answer:

The amplitude [tex]A(5) = 1 \ m[/tex]

Explanation:

From the question we are told that

     The  spring constant is  [tex]k = 100 \ N/m[/tex]

      The  damping constant is  [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]

       The mass is  [tex]m = 0.050 \ kg[/tex]

       The  maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           [tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]

substituting values

         [tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]

         [tex]A(5) = 1 \ m[/tex]

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

Answer:

The direction of the force is towards the East.

Explanation:

Using the right hand rule, the force on the current carrying conductor is east.

In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.

In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.

Answer:

metal conducts heat better than wood.

Explanation:

Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.

Answer:

Will be equal to alpha x r; less than UsN

Answer:

24.34 m/s

Explanation:

recall that one of the equations of motions takes the form:

v = u + at

where,

v = final velocity

u = initial velocity (given as 22.2 m/s)

a = acceleration (given as 2.68m/s²)

t = time elapsed during acceleration (given as 0.80s)

since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:

v = u + at

v = 22.2 + (2.68) (0.8)

v = 24.34 m/s

Answer:

Velocity of wave (V) = 5.2 × 10² m/s

Explanation:

Given:

Per unit length mass (U) = 4.80 × 10⁻³ kg/m

Tension (T)= 1,300 N

Find:

Velocity of wave (V)

Computation:

Velocity of wave (V) = √T / U

Velocity of wave (V) = √1300 / 4.80 × 10⁻³

Velocity of wave (V) = √ 270.84 × 10³

Velocity of wave (V) = 5.2 × 10² m/s

Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer

Explanation:

Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.

Friction is the force that prevents one hard material from scooting or rolling over the other.

Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.

We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.

Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.

In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.

Thus, the correct option is B.

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Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

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

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]

  = [tex]\frac{10+7155}{2650}[/tex]

[tex]\frac{1}{f}[/tex]  = [tex]\frac{7165}{2650}[/tex]

⇒ f = [tex]\frac{2650}{7165}[/tex]

      = 0.37

The focal length of the eye is 0.37 cm.

Answer:

Explanation:

change in magnetic flux = .16 x .16 x .510 - 0

= .013056 weber .

rate of change of flux = change in flux / time

= .013056 / 40.5 x 10⁻³

= .32237

voltage induced = .32237 V

electrical energy dissipated = v² / R where v is voltage , R is resistance

= .32237² / 6.35

= 16.36 x 10⁻³ J .

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

Answer:

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample