Given that the charge of the toaster is q =700 C
The current of the toaster is I = 10 A
We have to find the time and number of electrons.
Time can be calculated by the formula,
[tex]t=\frac{q}{I}[/tex]Substituting the values, the time will be
[tex]\begin{gathered} t=\frac{700}{10} \\ =70\text{ s} \end{gathered}[/tex]The number of electrons can be calculated by the formula,
[tex]n=\frac{q}{e}[/tex]Here, n is the number of electrons
and e is the charge of the electron whose value is
[tex]1.6\text{ }\times10^{-19}\text{ C}[/tex]Substituting the values, the number of electrons will be
[tex]\begin{gathered} n\text{ = }\frac{700}{1.6\times10^{-19}} \\ =4.375\text{ }\times10^{-17} \end{gathered}[/tex]Which is negatively charged?A. protonB. nucleusC. electronD. neutron
Protons, electrons an neutrons are the particles that make up atoms.
Protons have a positive electric charge, electrons have a negative electric charge an neutrons are electrically neutral.
The nucleus of an atom is made of protons and neutrons, so, its electric charge is positive an proportional to th
Look at the diagram below. From the frame of reference of the person riding
scooter B, what is the velocity of scooter A?
Scooter A
8 km/hr east
12 km/hr west
Scooter B
OA. 20 km/hr west
B. 20 km/hr east
OC. 4 km/hr east
D. 4 km/hr west
Kinetic energy differs from potential energy inA. Kinetic energy can be created or destroyed, while potential energy can not be created and destroyedB. Kinetic energy can be converted into various forms of energy, whereas potential energy can only be transformed into heat energy.C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.D. Kinetic energy is stored energy that has the capacity to do work, and potential energy is the energy of motion.
The kinetic enerfy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]And the potential energy is given by:
[tex]U=mgh[/tex]Where:
v = Velocity
h = height
m = mass
g = gravitational accceleration of earth
As we can see kinetic energy is associated to the movement and the potential energy is associated to the location, therefore the answer is:
C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.
B. MULTIPLE CHOICE. Choose the letter of the best answer2. the equation to calculate the momentum isa. p = mgb. p = mvc. p = mghd. p = mt
Answer:
b. p = mv
Explanation:
The momentum is the mass in motion, so it is calculated as the mass times the velocity. It means that the equation to calculate the momentum is
p = mv
Where m is the mass and v is the velocity of the object.
So, the answer is
b. p = mv
part B:
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
The magnitude of the acceleration of the box is 9.65 m/s².
What is the net force of the box?
The net force on the box is calculated as follows;
F(net) = F - Ff
where;
F is the applied forceFf is the force of frictionF(net) = F - μmgcosθ
where;
μ is the coefficient of friction given as 0.3θ is the angle of inclination of the plane = 55⁰m is the mass of the box = 15 kgF(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.71 N
The magnitude of the acceleration of the box is calculated as;
a = F(net) / m
a = (144.71) / (15)
a = 9.65 m/s²
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What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored in the 2.50 µF capacitor?
The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.
What is the potential difference of the circuit?The potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
where;
C is capacitance of the capacitorV is the potential differenceFor a parallel circuit the voltage in the circuit is always the same.
The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
2U = CV²
V = √2U/C
V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)
V = 12 V
The equivalent capacitance of C1 and C2 is calculated as follows;
1/C = 1/C₁ + 1/C₂
1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)
1/C = 1,173,611.11
C = 1/1,173,611.11
C = 8.52 x 10⁻⁷ C
The total capacitance of the circuit is calculated as follows;
Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C
Ct = 3.35 x 10⁻⁶ C
The total energy of the circuit is calculated as follows;
U = ¹/₂CtV²
U = ¹/₂(3.35 x 10⁻⁶ )(12)²
U = 2.41 x 10⁻⁴ J
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A hammer is used to hit a nail into a block of wood. The hammer hits the nail with a speed of 8.0 m/s and then stops. The hammer is in contact with the nail for 0.0015 s.,hammer has mass 0.15 kg.Calculate the average force between the hammer and the nail.
800 Newtons
Explanation
The average force is the force exerted by a body moving at a defined rate of speed (velocity) for a defined period of time.
the average force is given by:
[tex]F=ma[/tex]and
[tex]a=\frac{\Delta v}{\Delta t}[/tex][tex]\begin{gathered} F_{average}=m\frac{\Delta v}{\Delta t} \\ where\text{ m is the mass of the objectt} \\ \Delta v\text{ is the change in velocity} \\ \Delta t=\text{ time} \end{gathered}[/tex]Step 1
a) Let
[tex]\begin{gathered} m=0.0015\text{ kg} \\ \Delta v=0.0015s \\ \Delta v=8\text{ }\frac{m}{s} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} F=0.15\text{ kg}\frac{0-8\frac{m}{s}}{0.0015\text{ s}} \\ F=-800\text{ N} \end{gathered}[/tex]the negative sign indicates the force is in the opposite way ( the force is exerted by the nail to the hammer), so the force is opposite to the direction of the movement
so, the answer is
800 Newtons
I hope this helps you
Two 4.587 cm by 4.587 cm plates that form a parallel-plate capacitor are charged to +/- 0.671 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.257 mm?
ANSWER:
3.6 x 10^6 N/C
STEP-BY-STEP EXPLANATION:
Given:
Charge (q) = 0.671 nC = 0.671 x 10^-9 C
Side (s) = 4.587 cm = 4.587 x 10^-3 m
Vacuum permittivity (ε0) = 8.85 x 10^-12 F/m
We can calculate the electric field using the following formula:
[tex]\begin{gathered} E=\frac{q}{ε_0\cdot A} \\ \\ \text{ We replacing:} \\ \\ E=\frac{0.671\cdot10^{-9}}{(8.85\cdot10^{-12})(4.587\cdot10^{-3})(4.587\cdot10^{-3})} \\ \\ E=\:3603477.12=3.6\cdot10^6\text{ N/C} \end{gathered}[/tex]The electric field is equal to 3.6 x 10^6 N/C
A blink of an eye is a time interval of about 150ms for an average adult. The closure portion of the blink takes only about 55ms. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of 16.6 degree. What is the value of the angular acceleration the eyelid undergoes while closing 2. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is 1.25 cm?
ANSWER:
STEP-BY-STEP EXPLANATION:
The first thing is to convert the time into a second, just like this:
[tex]t=55\text{ ms}\cdot\frac{1\text{ s}}{1000\text{ ms}}=0.055\text{ s}[/tex]Now, convert the angular displacement of the eyelid from degrees to rad:
[tex]\partial\theta=16.6\text{\degree}\cdot\frac{2\pi\text{ rad}}{360\text{\degree}}=0.29\text{ rad}[/tex]We can calculate the angular velocity, dividing the angular momentum by the time, like this:
[tex]w=\frac{0.29}{0.055}=5.27\text{ rad/s}[/tex]The angular acceleration is calculated by means of the quotient of the difference in angular velocity and time, like this:
[tex]a_w=\frac{\delta w}{\delta t}=\frac{5.27-0}{0.15-0.055}=55.47\text{ rad/s}^2[/tex]the tangential acceleration would be:
The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.60 m/s
at an angle of 50.0 ∘
above the horizontal, and aims for a beetle on a leaf 2.30 cm above the water's surface.
The maximum height reached by the water is 20.2 cm and it will dislodge the beetle.
What is the maximum height reached by the water?
The maximum height reached by the water squirted by the arch fish is calculated by applying the following kinematic equation.
H = (v² sin²θ) / 2g
where;
v is the speed of the waterθ is the angle of projection of the waterg is acceleration due to gravityH = (2.6² x (sin50)² ) / (2 x 9.8)
H = 0.202 m
H = 20.2 cm
Thus, the water squirted by the arch fish is dislodge the beetle.
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The complete question is below:
The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 2.60 m/s
at an angle of 50.0 ∘ above the horizontal, and aims for a beetle on a leaf 2.30 cm above the water's surface. Will the water squirted by the arch fish dislodge the beetle?
A Hydrogen atom is a low density hot gas will give out what type of spectrum?A. A uniform spectrum containing all colorsB. A series of emission lines with equal spaces between the colorsC. A series of emission lines spaced in mathematical sequenceD.a uniform spectrum crossed by numerous dark absorption lines
The emission of photons takes place when an electron from higher energy orbitals jumps to a lower energy orbital.
Therefore the light emitted will correspond to the energy difference between the orbitals.
When the atom emits the photons, they will have energy equal to the energy difference between the orbitals of the Hydrogen. Therefore the spectrum obtained by the hydrogen gas will contain only those lines which correspond to the energy difference of the orbitals.
Therefore the hydrogen will emit a spectrum that contains a series of emission lines spaced in a mathematical sequence.
Therefore the correct answer is option C.
9. A yo-yo is moving in a horizontal circle of radius R. the yo-yo has a mass of 0.250 kg has a speed of 9 m/s and experience this a centripetal force of 26.6 N what is the radius of the circle that the yo-yo is moving in?
ANSWER:
B. 0.761 meters
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 0.250 kg
centripetal force (Fc)= 26.6 N
Speed (v) = 9m/s
We have that the centripetal force can be calculated using the following formula:
[tex]F_c=\frac{m\cdot v^2}{r}[/tex]We substitute each value and solve for the radius, just like this:
[tex]\begin{gathered} r=\frac{m\cdot v^2}{F_c} \\ r=\frac{0.25\cdot9^2}{26.6} \\ r=\frac{0.25\cdot81}{26.6} \\ r=0.761\text{ m} \end{gathered}[/tex]The radius is equal to 0.761 meters
A train car with a mass of 10kg and speed of 10 m/s is traveling to the right. Another train car with a mass of 20kg is moving to the left at -40 m/s. After the collision, the 10 kg train car is now moving at -20 m/s and we need to find the Velocity of the 20 kg train car.
When two particles collide and the masses of the particles are given, as well as the initial and final velocity of one particle and one of the velocities of the second particle, then the remaining velocity of the second particle is given by the expression:
[tex]v_2^{\prime}=\frac{m_1v_1+m_2v_2-m_1v_1}{m_2}[/tex]Which can be deduced from the Law of Conservation of Linear Momentum.
In the given problem, the initial and final velocities of the train car with mass 10kg are given, as well as the initial velocity of the 20kg car:
[tex]\begin{gathered} m_1=10kg \\ v_1=10\frac{m}{s} \\ v_1^{\prime}=-20\frac{m}{s} \\ \\ m_2=20kg \\ v_2=-40\frac{m}{s} \\ v_2^{\prime}=\text{ unknown} \end{gathered}[/tex]Replace those values into the given equation to find v₂':
[tex]\begin{gathered} v_2^{\prime}=\frac{(10kg)(10\frac{m}{s})+(20kg)(-40\frac{m}{s})-(10kg)(-20\frac{m}{s})}{20kg} \\ \\ \Rightarrow v_2^{\prime}=-25\frac{m}{s} \end{gathered}[/tex]Therefore, the velocity of the 20kg train car after the collision, is: -25 m/s.
A ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22.0%,
If a ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22°, and the refractive index of the liquid would be 1.334.
What is refraction?It is the phenomenon of bending of light when it travels from one medium to another medium. The bending towards or away from the normal depends upon the medium of travel as well as the refractive index of the material.
By using Snell's law,
Refractive index of the liquid = sin(i) /sin(r)
=sin(30) /sin(22)
= 1.334
Thus, the refractive index of the liquid would be 1.334.
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why is the hubble space telescope in space instead of on the ground?
Answer:EARTH'S ATMOSPHERE ALTERS AND BLOCKS THE LIGHT THAT COMES FROM SPANCE.
Explanation:
Hubble orbits above Earth's atmosphere, which gives it a better view of the universe than telescopes have at ground level.
A golf ball is initially on a tee when it is
struck by a golfer. The ball is given an
initial velocity of 50 m/s at a 37° angle. The
ball hits the side of a building that is 200
meters horizontally away from the golfer.
(a) What are the horizontal and vertical
components of the ball's initial
velocity?
(b) How much time elapses before the
ball strikes the side of the building?
(c) How far from the ground does the ball
strike the building?
Answer:
a.
[tex]horizontal=39.9[/tex] m/s
[tex]vertical=30.1[/tex] m/s
b.
[tex]t=5.009[/tex]
c.
[tex]y=27.7[/tex]
Explanation:
Lets write down what we were given.
Angle = 37°
Initial Velocity = 50 m/s
Displacement in x direction = 200 m
Take note:
I am having some trouble with the theta symbol so let theta = [tex]N[/tex]
Lets do question C first.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
Rearranging terms, we have
[tex]y=(tan(N)*x)-[\frac{g}{2(v_{0}cos(N))^{2} } ]x^{2}[/tex]
Now lets substitute our numbers in for the variables. Then simplify.
[tex]y=(tan37*200)-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[0.0030761]200^{2}[/tex]
[tex]y=150.7108-(0.0030761*40000)[/tex]
[tex]y=150.7108-123.0444[/tex]
[tex]y=27.7[/tex]
Now lets do question B.
Lets steal this from the last question.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
We can substitute [tex]t[/tex] for [tex]\frac{x}{v_{0}cos(N) }[/tex]
[tex]y=(v_{0}sin(N))*(t)-\frac{1}{2}g(t) ^{2}[/tex]
We can rewrite the equation as
[tex](v_{0}sin(N)(t)-\frac{1}{2}*(g(t)^{2})=y[/tex]
Now lets substitute our numbers in for the variables.
[tex](50sin(37)(t)-\frac{1}{2}*(9.81(t)^{2})=27.7[/tex]
After some painful algebra and factoring we get
[tex]30.09075115t-4.905t^{2}=27.6664[/tex]
Subtract [tex]27.6664[/tex] from both sides.
[tex]30.09075115t-4.905t^{2}-27.6664=0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
After some more painful algebra we get
[tex]t=5.00854263, 1.12616708[/tex]
1.126 does not make any sense so.
[tex]t=5.009[/tex]
Finally lets do question A.
Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.
Let [tex]A=horizontal[/tex] and [tex]O=vertical[/tex]
[tex]cos(37)=\frac{A}{50}[/tex]
[tex]A=39.9[/tex]
[tex]sin37=\frac{O}{50}[/tex]
[tex]O=30.1[/tex]
The couple required to hold a triple turn of 1.5cm² area in equilibrium when carrying a current 2A at 70° to a field with 0.15T is?
The couple or torque required to hold the triple turn is 1.27 x 10⁻⁴ Nm.
What is the couple or torque required?
The couple required to hold the triple turn is calculated as follows;
τ = M x Bsinθ
where;
M is the magnetic moment B is the magnetic field strengthThe magnetic moment is calculated as follows;
M = NIA
where;
N is number of turns = 3I is current = 2 AA is the area of the loop = 1.5 cm² = 0.00015 m²M = (3) x (2) x (0.00015)
M = 0.0009 m²A
The torque or couple required is calculated as;
τ = (0.0009) x (0.15 x sin70)
τ = 1.27 x 10⁻⁴ Nm
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itial height at Two balls are thrown vertically from the same • Ball I is launched upward with an initial velocity voj = + 10m/s. Ball 2 is launched downward with an initial velocity vo2 = - 10m/s. same The distance between the two balls after I second from the beginning of motion is:
Given
vo1 = +10 m/s
vo2 = -10 m/s
Procedure
Using the free fall equations, we have:
[tex]\begin{gathered} x1=v_{o1}t-\frac{1}{2}gt^2 \\ x1=10*1-\frac{1}{2}9.8*1 \\ x1=5.1m \end{gathered}[/tex][tex]\begin{gathered} x2=v_{o2}t-\frac{1}{2}gt^2 \\ x2=-10*1-\frac{1}{2}9.8*1 \\ x2=-14.9m \end{gathered}[/tex][tex]\begin{gathered} x1-x2=5.1-\lparen-14.9) \\ x1-x2=20 \end{gathered}[/tex]The distance between the balls would be 20m
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the objects speed were halved in the mass was tripled, what would happen to the centripetal force?
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the object's speed were halved in the mass was tripled, then the centripetal force would be 0.75 times the original centripetal force.
What is a uniform circular motion?It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion.
A mass m object travels in a circle at a constant speed v. The object's centripetal force is F. The centripetal force would be 0.75 times greater if the object's mass were tripled and its speed was cut in half.
Centripetal force = m × v²/r
=3m × (0.5v)² / r
= 0.75 mv² / r
Thus, the centripetal force would become 0.75 times the original centripetal force.
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All matter is composed of quarks and leptons. Is this true or false?
All matters are made up of protons, neutrons and electrons. And protons and neutrons are made up of fundamental particles known as quarks. Leptons are fundamental particles with half-integer spin. An electron is a lepton.
Thus the given statement is true.
A 4000-kg truck traveling with a velocity of 20 m/s due south collides head-on with a 1320- kg car traveling with a velocity of 10 m/s due north. The two vehicles stick together after the collision.A. What are the magnitude and the direction of the momentum of each vehicles Prior to the collision?B. What are the magnitude and the direction of the velocity of both vehicles after their collide?
Before we begin, we will establish that the north direction is the positive direction which means that the south direction is negative.
A.
The momentum of an object is given by:
[tex]p=mv[/tex]For the truck we know that the velocity is -20 m/s (since it is traveling south) and its mass is 4000 kg, then its momentum is:
[tex]p=(4000)(-20)=-80000[/tex]Therefore, the momentum of the truck is -80000 kg m/s; this means that its magnitude is 80000 kg m/s and its direction is south.
For the car we know that the velocity is 10 m/s and its mass is 1320 kg, then its momentum is:
[tex]p=(1320)(10)=13200[/tex]Therefore, the momentum of the car is 13200 kg m/s which means that its magnitude is 13200 kg m/s and its direction is north.
B.
In a collision the momentum is conserved, that is, the total initial and final momentum is equal, that is:
[tex]p_i=p_f[/tex]In this case, we know that the vehicles stick together after they collide, then we have:
[tex]m_tv_t+m_cv_c=(m_t+m_c)u[/tex]where u is the velocity of the vehicles after they collide. Plugging the values we know, we have that:
[tex]\begin{gathered} -80000+13200=(4000+1320)u \\ u=\frac{-80000+13200}{4000+1320} \\ u=-12.56 \end{gathered}[/tex]Therefore, the final velocity of the system is -12.56 m/s which means that the magnitude of the velocity is 12.56 m/s and its direction is south.
Find the magnitude of the sumof these two vectors:B63.5 m101 m57.0°
Vector diagram:
The resultant vector is given as,
[tex]R=\sqrt[]{A^2+B^2+2AB\cos \theta}[/tex]Here, θ is the angle between vector A and B.
Substituting all known values,
[tex]\begin{gathered} R=\sqrt[]{(63.5)^2+(101)^2+2\times101\times63.5\times\cos (33^{\circ})} \\ =158.08\text{ m} \end{gathered}[/tex]Therefore, the resultant magnitue of the sum of these two vectors are 158.08 m.
The x-component of the magnitude is given as,
[tex]\begin{gathered} R_x=101\cos (57^{\circ})+63.5\cos (90^{\circ}) \\ =55.0\text{ m} \end{gathered}[/tex]The y- component of the magnitude is given as,
[tex]\begin{gathered} R_y=63.5\sin (90^{\circ})+101\sin (57^{\circ}) \\ =148.2\text{ m} \end{gathered}[/tex]Therefore, the direction is given as,
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{R_y}{R_x}) \\ =\tan ^{-1}(\frac{148.2\text{ m}}{55.0\text{ m}}) \\ =69.63^{\circ} \end{gathered}[/tex]Therefore, the direction of the resultant vector is 69.63°.
The diagram below represents a 2.0 kg toy car moving at across the speed of 3.0 meters per 2nd counter clockwise in a circular path with a radius of 2.0 meters.At the Instant shown in the diagram, the direction of the centripetal force acting on the car is_____.
Given data
The mass of the toy car is m = 2 kg
The speed of the car is v = 3 m/s
The radius of the circular track is r = 2 m
The centripetal force is always in the same direction as that of the centripetal acceleration.
The centripetal acceleration direction is towards the center of the circle, towards west.
Therefore, the direction of the centripetal force points to the west direction.
Thus, the direction of the centripetal force at this instant is towards the west.
i’m still really confused on how to actually calculate it
Question 6:
Given information:
Distance travelled by bus,
[tex]s=10100\text{ m}[/tex]Average velocity of the bus,
[tex]v=5.6\text{ m/s}[/tex]We need to find the time taken by bus to reach school. Let t be the time taken by bus to reach school. The velocity of the bus is given as,
[tex]v=\frac{s}{t}[/tex]The expression for the time is given as,
[tex]t=\frac{s}{v}[/tex]Substituting all known values,
[tex]\begin{gathered} t=\frac{10100\text{ m}}{5.6\text{ m/s}} \\ \approx1804\text{ s} \\ \approx30\text{ min 4 sec} \end{gathered}[/tex]Therefore, the bus required 30 min 4 sec to reach school.
It takes 5 seconds for a 2 kg box to be pushed 10 meters from rest. What was the forceof the push?
Given data:
* The mass of the box is 2 kg.
* The time taken by the box to travel the given distance is 5 seconds.
* The distance traveled by the box is 10 meters.
* The initial velocity of the box is 0 m/s.
Solution:
By the kinematics equation, the distance traveled by the box in terms of its acceleration is,
[tex]S=ut+\frac{1}{2}at^2[/tex]where u is the initial velocity, t is the time taken, a is the acceleration, and S is the distance traveled,
Substituting the known values,
[tex]\begin{gathered} 10=0+\frac{1}{2}\times a\times(5)^2 \\ 10=\frac{25}{2}\times a \\ a=10\times\frac{2}{25} \\ a=0.8ms^{-2} \end{gathered}[/tex]By the Newton's second law, the force exerted on the box in terms of the acceleration is,
[tex]F=ma[/tex]where m is the mass of the box, a is the acceleration and F is the force,
Substituting the known values,
[tex]\begin{gathered} F=2\times0.8 \\ F=1.6\text{ N} \end{gathered}[/tex]Thus, the force of the push is 1.6 N.
Determine the resistance, in milliOhms, of a metal rod 2.96 m long, 0.89cm diameter and composed of aluminum of resistivity 2.8 x 10-8 Ωm .
The resistance R of a rod with length L, cross-sectional area A and resistivity ρ is given by:
[tex]R=\frac{\rho L}{A}[/tex]On the other hand, the area of a circle with diameter D is given by:
[tex]A=\frac{\pi}{4}D^2[/tex]Then, the resistivity of the rod in terms of its diameter is:
[tex]R=\frac{4\rho L}{\pi D^2}[/tex]Replace L=2.96m, D=0.89cm and ρ=2.8×10^(-8)Ωm to find the resistance of the metal rod:
[tex]\begin{gathered} R=\frac{4\rho L}{\pi D^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89cm)^2} \\ \\ =\frac{4(2.8\times10^{-8}\Omega m)(2.96m)}{\pi(0.89\times10^{-2}m)^2} \\ \\ =1.332232...\times10^{-3}\Omega \\ \\ \approx1.33m\Omega \end{gathered}[/tex]Therefore, the resistance of the metal rod is approximately 1.33 miliOhms.
Vector A= 30 m/s towards East and vector B= 80 m/s towards south. Find A- B [Perform the subtraction of the vector].
ANSWER
[tex]\begin{equation*} 85.44\text{ }m\/s \end{equation*}[/tex]EXPLANATION
First, let us make a sketch of the two vectors:
The vector A - B is represented by line BA in the figure above.
To evaluate A - B, we will apply the Pythagoras theorem:
[tex]\begin{gathered} A-B=BA=\sqrt{(30)^2+(80)^2} \\ A-B=BA=\sqrt{900+6400} \\ A-B=BA=\sqrt{7300} \\ A-B=BA=85.44\text{ }m\/s \end{gathered}[/tex]That is the answer.
carts, bricks, and bands
7. Which of the following conclusions are supported by the data in Table 2?
a. Adding bricks to a cart has no affect upon the cart's acceleration.
b. Increasing the mass of an object causes a decrease in its acceleration.
c. An increase in the number of rubber bands causes an increase in the acceleration.
d. The more mass that an object has, the more acceleration that it will acquire when pushed.
The conclusions that are specifically supported by the data in Table 2 Increasing the mass of an object causes a decrease in its acceleration. That is option B.
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 5 ----> no bricks = 0.99 m/s²
Trial 6 ----> cart with one brick = 0.50 m/s²
Trial 7 ----> cart with two bricks = 0.32 m/s²
Trial 8 -----> cart with three bricks = 0.25 m/s²
From the information above, progressive increase the the quantity and mass of the bricks lead to a decrease in the acceleration of the cart with a constant force from only 2 bands.
This occurred because the mass is inversely proportional to acceleration.
Learn more about acceleration here:
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i am not sure the best way to solve this problem
ANSWER
14.11 s
EXPLANATION
We know that in total, the runner will run a distance of 100m. He runs at constant acceleration for a while and then his velocity gets constant until the end of the track - this means that in the last part, his acceleration is zero.
So we have two parts:
For the first part, we have the acceleration and time. If we set that the initial position is zero, as shown in the diagram above, and that the runner starts from rest - therefore, his initial velocity is zero - we can find the distance of the first part of the path, which we'll call x1:
[tex]x_1=x_0+v_0t+\frac{1}{2}at^2[/tex]Since x0 and v0 are both zero, then those terms get cancelled:
[tex]x_1=\frac{1}{2}\cdot a\cdot t^2=\frac{1}{2}\cdot1.5\cdot6^2=27m[/tex]So the first part of the track, where the runner is speeding up, has a distance of 27m. Therefore, the rest of the track where the runner runs at constant acceleration is:
[tex]100-27=73[/tex]73m.
We want to find the time it took the runner to run the whole 100m. We know that he did the first part in 6 seconds. To find the time of the second part, we can use the distance we just found. Let's call it xf:
[tex]x_f-x_1=\frac{1}{2}at^2+v_0t[/tex]We know that the acceleration in this part of the track is zero and the initial velocity for this part is the velocity the runner had when he reached 6 seconds - i.e. 27m:
[tex]73m=v_1\cdot t[/tex]We don't know the time and we don't know the velocity, but we can find the second one using the formula for velocity for the first part of the track with t = 6s:
[tex]\begin{gathered} v_1=a\cdot t+v_0 \\ v_1=1.5\cdot6 \\ v_1=9m/s \end{gathered}[/tex]Now we can find the time for the second part of the track:
[tex]\begin{gathered} 73m=9m/s\cdot t \\ t=\frac{73m}{9m/s} \\ t\approx8.11s \end{gathered}[/tex]Therefore, the total time it took the runner to run 100m was:
[tex]\begin{gathered} t=6s+8.11s \\ t=14.11s \end{gathered}[/tex]14.11 s
An object has an excess charge of −1.6 × 10−17 C. How many excess electrons does it have?
Given:
The charge on the object is
[tex]q=-1.6\times10^{-17}\text{ C}[/tex]Required: Number of electrons
Explanation:
The number of electrons can be calculated using the quantization of charge
[tex]q\text{ = ne}[/tex]Here, n is the number of electrons
e is the charge on the electron whose value is
[tex]e\text{ = -1.6}\times10^{-19}\text{ C}[/tex]
On substituting the values, the number of electrons will be
[tex]\begin{gathered} n=\frac{q}{e} \\ =\frac{-1.6\times10^{-17}}{-1.6\times10^{-19}} \\ =100 \end{gathered}[/tex]Final Answer: The object has an excess of 100 electrons.