Physical science section 1 review chapter 4 newtons laws
I can give more points if you answer it

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ([tex]a_c[/tex]) is given by:

[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

Answer:

in this video waves are coming up for the BOTTOM to the top of the sandbar

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Answer:

(A) The time taken for the ball to stop is 8.7 x 10⁻³ s

(B) The distance traveled by the baseball before stopping is 0.3 m

Explanation:

Given;

mass of the baseball, m = 0.14 kg

velocity of the baseball, v = 23 m/s

force exerted on the baseball by the catcher, F = 370 N

(A) The time taken for the ball to stop;

[tex]F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.14 \times 23}{370} \\\\t = 8.7\times 10^{-3} \ s\\\\t = 8.7 \ ms[/tex]

(B) The distance traveled by the baseball before stopping is calculated as;

acceleration of the ball, [tex]a = \frac{v}{t} = \frac{23}{8.7\times 10^{-3}} = 2643.678 \ m/s^2[/tex]

Distance traveled, s;

s = ut + ¹/₂at²

s = (23)(8.7 x 10⁻³) + ¹/₂(2643.678)(8.7 x 10⁻³)²

s = 0.2001 + 0.1001

s = 0.3 m

Answer:

Impulse of force = 300Ns

Explanation:

Given the following data;

Mass = 25kg

Change in velocity = 12m/s

To find the impulse;

Impulse is given by the formula;

[tex] Impulse \; of \; force = mass * change \; in \; velocity [/tex]

Substituting into the equation, we have;

[tex] Impulse \; of \; force = 25 * 12 [/tex]

Impulse of force = 300Ns

Answer:

Explanation:

The vertical component of the acceleration of a sailplane is zero , that means the sailplane is experiencing net force of zero in vertical direction . Its weight is acting in downward direction . So airplane is also experiencing an upward force equal to its weight  which is making net force equal to zero on it . This force is given as 5.20 k N .

So sailplane is experiencing an upward force equal to its weight . This force is generated due to air pushing up against its wings .

We know that every force generated has equal and opposite reaction force . Air is generating force on wings of sailplane , hence wings will also exert equal force on air on downward direction . This force will be transmitted to the earth by air .

Hence the gravitational force on Earth due to the sailplane will be equal to weight of sailplane . This force is 5.2 kN .

._.Answer:.

ijji._. ji

Explanation:.

The correct options are (a &b) .

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Answer: A & B

Explanation: just took it !

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

[tex]F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\[/tex]

[tex]But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz[/tex]

Answer:

9.42 m/s

Explanation:

a) Using Newton's law of motion formula:

[tex]\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s\\[/tex]

The speed of the discus at release (v) is:

v = ωr; where r = radius of discus

diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m

v = ωr = 10.47 * 0.9

v = 9.42 m/s

Answer:

5 atoms of Boron

Explanation:

Boron makes up approximately 15.944% of the mass and the rest of the 84.056% is Fluorine. There is 5 Atoms because Boron atomic mass is 10.811 in 1 molecule of BF3 and you wanted 5 Molecules.

To see the number of atoms of an element in a given molecule we need to multiply stoichiometry to the number that is written on the foot of the element that is stoichiometry. Therefore, in 5 molecules of  BF[tex]_3[/tex], the number of boron atoms is 5.

Atom is the smallest particle of any element, molecule or compound. Atom can not be further divided. Atoms contains nucleus in its center and electron that revolve around the atom in fixed orbit.

In the nucleus, proton and neutron are present. Electron has -1 charge while proton has +1 charge. Neutron is neutral that is it has no charge. So overall the charge of nucleus is due to only proton, not by neutron.

In one molecule of  BF[tex]_3[/tex], the number of boron atoms is 1. So, in 5 molecules of   BF[tex]_3[/tex], the number of boron atoms is 5.

Therefore, in 5 molecules of   BF[tex]_3[/tex], the number of boron atoms is 5.

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Answer:

c. A number and a unit

Explanation:

A scalar is a real number. We often use the term scalar in the context of vectors or matrices, to stress that a variable such as a is just a real number and not a vector or matrix.

Answer:

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

(a) The maximum force on the  electron due to the magnetic field will be  F= 79.02 x 10⁻¹³ N .

(b) Minimum force will be zero when electron moves along the direction of magnetic field .

Whenever a current is passes through a wire then the magnetic fields are generated around the wire and if any other charged particle comes under the influence of this magnetic field then the magnetic force is applied in the charge.

Let v be the velocity acquired by an electron in electric field

[tex]Vq=\dfrac{1}{2}mv^2[/tex]

V is potential difference applied on charge q ,

m is mass of charge ,

v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on the moving electron when it moves perpendicular to the magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on an electron and v is velocity of electron

F=Bqv

F= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

F= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

Hence the maximum force on the  electron due to the magnetic field will be  F= 79.02 x 10⁻¹³ N and the Minimum force will be zero when electron moves along the direction of magnetic field .

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Answer:

a

Explanation:

AAAA

Answer:

  f ’= 3019.8 Hz

Explanation:

This is an exercise of the Doppler effect, that is, of the relative movement of the source (s) and the observer (o)

         [tex]f' = f_o \frac{v +v_o}{v -v_s}[/tex]

in the case of the source and the observer approaching, if they move away the signs are exchanged

let's calculate

       f ’= 2500 [tex]\frac{343 + 23}{343 - 40}[/tex]

       f ’= 2500 [tex]\frac{366}{303}[/tex]

       f ’= 3019.8 Hz

Answer:

Thunderbird has fallen behind the Mercedes Benz by 1017.49 m

Explanation:

Given the data in question;

initial speed of the ford u1 = 73.5 m/s

distance d1 = 250 m

t1 = 5.00 s

d2 = 400 m

Now, let the time taken to stop be t2 and deceleration is a1

so,

a1 = u1² / (2 × d1)

a1 = (73.5)² / (2 × 250)

a1 = 10.8045 m/s²

Now , for acceleration is a2

a2 = v² / (2 × d2)

a2 = (73.5)² / (2 × 400)

a2 = 6.7528 m/s²

total time spend = 5 + u/a1 + u/a2

total time spend = 5 + (73.5/10.8045) + (73.5/6.7528)

total time spend = 22.687 sec

Now, distance Mercedes is ahead = 22.687 × 73.5 - 400 - 250

= 1667.4945 - 400 - 250

= 1017.49 m

Therefore, Thunderbird has fallen behind the Mercedes Benz by 1017.49 m

Answer:

C. The star is receding from you with the speed of 3000 km/s

Explanation:

To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.

We solve for V

V = 3x10⁶m/s

V = 3000km/s

We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.

Thank you!

Answer:

a) d= 1393 m

b) θ= 21º S of E.

Explanation:

a)

       [tex]d = \sqrt{(1300m)^{2} + (500m)^{2} } = 1393 m (1)[/tex]

b)

       [tex]tg \theta = \frac{500m}{1300m} = 0.385 (2)[/tex]

       ⇒ θ= tg⁻¹ (0.385) = 21º S of E.

Answer:

Shadows are formed when an opaque object or material is placed in the path of rays of light. The opaque material does not let the light pass through it. The light rays that go past the edges of the material make an outline for the shadow.

Explanation:

Hope it is helpful.....

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ([tex]v[/tex]), measured in meters per second, by using this kinematic equation:

[tex]v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}[/tex] (1)

Where:

[tex]a[/tex] - Acceleration, measured in meters per square second.

[tex]\Delta s[/tex] - Travelled distance, measured in meters.

[tex]v_{o}[/tex] - Initial velocity, measured in meters per second.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 2.35\,\frac{m}{s^{2}}[/tex] and [tex]\Delta s = 595\,m[/tex], the final velocity of the rocket is:

[tex]v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}[/tex]

[tex]v\approx 52.882\,\frac{m}{s}[/tex]

The time associated with this launch ([tex]t[/tex]), measured in seconds, is:

[tex]t = \frac{v-v_{o}}{a}[/tex]

[tex]t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }[/tex]

[tex]t = 22.503\,s[/tex]

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

[tex]v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})[/tex] (2)

Where:

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]v[/tex] - Final speed, measured in meters per second.

[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]s_{o}[/tex] - Initial height, measured in meters.

[tex]s[/tex] - Final height, measured in meters.

If we know that [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex], [tex]a = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]s_{o} = 595\,m[/tex], then the maximum height reached by the rocket is:

[tex]v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})[/tex]

[tex]s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]

[tex]s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]

[tex]s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]s = 737.577\,m[/tex]

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

[tex]s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}[/tex] (2)

Where:

[tex]s_{o}[/tex] - Initial height, measured in meters.

[tex]s[/tex] - Final height, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]s_{o} = 595\,m[/tex], [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]s = 0\,m[/tex] and [tex]a = -9.807\,\frac{m}{s^{2}}[/tex], then the time needed by the rocket is:

[tex]0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]

[tex]-4.904\cdot t^{2}+52.882\cdot t +595 = 0[/tex]

Then, we solve this polynomial by Quadratic Formula:

[tex]t_{1}\approx 17.655\,s[/tex], [tex]t_{2} \approx -6.872\,s[/tex]

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

Answer:

case A) tau_net = -243.36 N m, case B)    tau_net = 783.36 N / m,      tau_net = -63.36 N m,  case C)  tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

         Σ τ  = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

         tau_net = m g x - F x2

          tau_nett = -28.8 9.8 1.5 + 180 1

         tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

 the force is up

          tau_nett = -28.8 9.8 1.5 - 180 2

          tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

        tau_nett = -28.8 9.8 1.5 + 180 2

         tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

 tau_nett = -28.8 9.8 1.5 - 180 3

 tau _net = - 963.36 N m

3 counterclockwise

       tau_nett = -28.8 9.8 1.5 + 180 3

       tau_net = 116.64 Nm

the sitam rotated counterclockwise

Answer:

it spins

Explanation:

like a lot, bro i was on one before and it was pretty nice :)

Answer:

Current through them, voltage across them.

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

[tex] V = IR[/tex]

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

A resistor can be defined as an electronic component that opposes the free flow of current from one point to another in an electrical circuit.

In a series circuit of two resistors, the resistors have the same current through them; in a parallel circuit of two resistors, the resistors have the same voltage across them.

Answer: m/s

Explanation:

The slope is the change in y over the change in x. If the graph is velocity over time, then the units will be meters per second

Answer:

θ₂ = 40.5º

Explanation:

For this exercise we must use the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for the incident ray and index 2 is for the refracted ray

in this case the incident ray has an angle of θ₁ = 60º and the refractive index of the water is

          n₂ = 1,333

        sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]

let's calculate

        sin  θ₂ = 1 / 1.3333 sin 60

        sin θ₂ = 0.64968

        θ₂ = sin⁻¹ (0.64968)

        θ₂ = 40.5º

Answer:

a) The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.

b) the net electric flux is  -821.09 N.m²/c

c) the net charge inside the cylinder is -7.2666 nC

Explanation:

Given the data in the question;

a) (a) What type of charge distribution is inside the surface?

Based on the Image of the question below;

The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.

b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface?

we know that; Electric flux is;

∅ = -2 × E × Area

we substitute

∅ = -2 × 300 × π ( 0.66 m )²

= -821.09 N.m²/c

Therefore; the net electric flux is  -821.09 N.m²/c

c) What is the net charge inside the cylinder? nC

from Gause's law;

∅ = q/∈₀

q = ∅ ∈₀

we know permittivity ∈₀ = 8.85 × 10⁻¹²)

we substitute

q = -821.09 N.m²/c  × (8.85 × 10⁻¹²)

q =  -7.2666 × 10⁻⁹ C

q = -7.2666 nC

Therefore, the net charge inside the cylinder is -7.2666 nC

Answer:

Explanation:

d = rt only works for calculating average speeds.

100 = r(140)

r = 0.714 m/s

Answer: 0.71m/s

Explanation:

Formula: Speed = Distance/time

Speed = 100m/140s

Speed = 100/140 / 20/20

Speed = 5/7 or 0.71m/s

Answer:

The heavier the object and the higher it is above the ground, the more gravitational potential energy it holds. Gravitational potential energy increases as weight and height increases. Potential energy is energy that is stored in an object or substance.