The sign (+ or −) and charge of the projectile, if the magnetic force on the projectile is 3.22 10-10 N, directed due west is mathematically given as
q=+1.11524613*10^{−4}C
What is the sign (+ or −) and charge?Generally, the equation for Force is mathematically given as
[tex]F=Bqvsin\theta[/tex]
Therefore
[tex]3.22 10^{-10}= 5.88 10^{-5 } *633*sin 56*q\\\\q=\frac{3.22 10^{-10}}{5.88 10^{-5 } *633*sin 56}[/tex]
q=+1.11524613*10^{−4}C
In conclusion, the sign (+ or −) and charge
q=+1.11524613*10^{−4}C
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Question Completion Status: 1 points QUESTION 1 The following excerpt is taken from the Governors Highway Safety Association (GHSA) website (https://www.ghsa.org): Setting speed limits has traditionally been the responsibility of states, except for the period of 1973-1994. During that time, the federal government enacted mandatory speed limit ceilings on interstate highways and similar limited access roads through a National Maximum Speed Limit. ✓ Sav Congress repealed the National Maximum Speed Limit in 1995. Since then, 41 states have raised speed limits to 70 mph or higher on some portion of their roadway systems. In many states, maximum speeds vary depending on vehicle type (car or truck), roadway location (urban or rural), or time of day. GHSA tracks state maximum speed limits for both urban and rural interstates, as well as other limited access roads. In a few states, speed limits are not set by law. Your textbook includes the following physics problem: If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time. To complete the Civic Engagement component of this course, write an essay about the National Speed Limit. In this essay, do the following: 1. Solve the above physics problem. 2. Use your solution to the above physics problem and the information contained in the above excerpt to argue for the reinstatement of a National Maximum Speed Limit. Your orrny should be 3000 to 5000 characters long (not including spaces) You are expected to format all variables/equations using the math editor, which you access
There should be a reinstatement of the National Maximum Speed Limit for the safety of road users because the minimum braking distance is 2,25m.
Calculations and ParametersHence, we can see that the minimum braking distance for the car would be:
d'/d= (v'1/v1)^2 = (1.5/1)^2
= 2.25m
The speed limit ensures that drivers do not exceed a set speed limit in order to reduce road accidents and keep road users safe and also the pedestrians would be able to navigate easier.
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the sound of a lamb is ....... due to its pitch depending upon its ....... frequency?
The sound of a lamb is grave due to its high pitch depending on its low frequency. Option B is correct.
What is the frequency of the sound?A sound pressure wave's frequency is the number of times it repeats itself every second.
The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.
Pitch and the frequency of sound are inverse to each other. A lamb's sound is grave because of its high pitch and low frequency.
The sound of a lamb is grave due to its high pitch depending on its low frequency.
Hence, option B is correct.
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A rod of length 58.0 cm and mass 1.10 kg is suspended by two strings which are 35.0 cm long, one at each end of the rod.
The string on side B is cut. Find the magnitude of the initial acceleration of end B.
The string on side B is retied and now has only half the length of the string on side A.
Find the magnitude of the initial acceleration of the end B when the string is cut.
a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω
b)ω = [tex]\frac{3g}{2l}[/tex]
c)14.7 the magnitude of the initial acceleration of the end B when the string is cut.
What is Acceleration?Acceleration is the rate of change of the velocity of an object with respect to time.
Given
A rod of length 56.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 cm long, one at each end of the rod.
When the string is cut rod rotates about other end.
Let's take momentum equation about that end:
mg*[tex]\frac{1}{2}[/tex] = Iω [I = [tex]ml^\frac{2}{3}[/tex]]
a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω
Angular acceleration:
b)ω = [tex]\frac{3g}{2l}[/tex]
c)Acceleration of end B:
a = lω= [tex]\frac{3g}{2}[/tex] =[tex]\frac{3*9.8}{2}[/tex]
a = 14.7 [g=9.8]
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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
The inner radius of the yoyo is r = 2.14 cm, and the outer radius is R = 4.00 cm, and the moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is I[tex]_{cm}[/tex] = 1.01×10-4 kgm2.
1. Determine the linear acceleration of the yoyo.
2. Determine the angular acceleration of the yoyo.
3. What is the weight of the yoyo? (Hint: It's not 150 g)
4. What is the tension in the rope?
5. If a 1.27 m long section of the rope unwinds from the yoyo, then what will be the angular speed of the yoyo?
(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
Linear acceleration of the yoyoThe linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
I is moment of inertiaα is angular accelerationT is tension in the roper is inner radiusR is outer radiusf is frictional forcerT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
a is the linear acceleration of the yoyoTorque equation for frictional force;
[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]
solve (1) and (2)
[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]
since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]
Angular acceleration of the yoyoα = a/R
α = 3.21/0.04
α = 80.25 rad/s²
Weight of the yoyoW = mg
W = 0.15 x 9.8 = 1.47 N
Tension in the ropeT = mg = 1.47 N
Angular speed of the yoyov² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
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importance of states of matter
Answer:
STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.
Explanation:
A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp..
a) 54.6%
b) 32%
c) 46.4%
d) 54.65%
A. The efficiency of the ramp is 54.6 %.
Velocity ratio of the rampV.R = distance moved by effort/distance moved by load = L/h
V.R = (6.12)/(1.53) = 4
Mechanical advantage of the rampM.A = Load/Effort
M.A = (451 x 9.8)/(2025)
M.A = 2.183
Efficiency of the rampE = (M.A / V.R) x 100%
E = (2.183 / 4) x 100%
E = 54.6 %
Thus, the efficiency of the ramp is 54.6 %.
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You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?
If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?
The work done is 8427 J while the energy expended is 80.25 W.
What is work done?Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;
W = mgh
Total mass of the water bucket and chain = 13.9 kg + 19.3 kg = 33.2Kg
Distance covered = 25.9 m
W = 33.2Kg * 9.8 m/s^2 * 25.9 m
W = 8427 J
Recall that the work done = Energy expended
Power = Energy expended/ Time
Power = 8427 J/ 1.75 * 60 seconds
Power = 80.25 W
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Theory of uniform accelerated motion lab reports
The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
What is the theory of uniform accelerated motion?The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.
This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.
In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
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In each case below, a charged rod, made of the dense rubber ebonite, comes close or is in contact with the top of an electroscope. The ball on top of the electroscope is directly connected to the two metal leaves suspended in the flask.
Which image represents a gaining of a charge on the leaves of the electroscope by conduction?
Answer:
(B) (the leaves will separate when subjected to a net charge)
The image that represents a gaining of a charge on the leaves of the electroscope by conduction is image B.
What is electroscope?An electroscope is a device that detects the presence of an electric charge or ionizing radiation by suspending a pair of thin gold leaves from an electrical conductor that leads to the outside of an insulating container.
When an electrical charge is brought close to or in contact with a conductor, the leaves in an electroscope spread apart.
This occurs because the same charge, whether positive or negative, is received by the leaves, causing their repulsion (spreading apart).
A negatively charged rod is brought in close proximity to an uncharged electroscope. As a result, the electroscope leaves become further apart.
Thus, image B is representing the given scenario.
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What is the size of the image on the retina of a size object 1.5 cm, placed at a distance (120 cm) away? Take the .lens-to-retina distance to be 2 cm
Answer: 0.025 cm
Explanation:
The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm
The magnification of the lens is given by
m = v/u = h'/h
Where h' is the height of the imagev/u = h'/ h
h' = ( v/u ) × h
h' = ( 2cm /-120cm ) × 1.5cm
h' = - 0.025cm
Therefore, the height of the image is 0.025 cm
A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.
a) What is the slowest speed that the rock can travel and still maintain a circular path?
b) What is the tension in the string at the bottom of the swing?
Hello!
a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.
At the top of the circle, we have:
- Force of gravity (downward)
*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.
Therefore, only the force of gravity produces the net centripetal force:
[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]
We can simplify and rearrange the equation to solve for 'v'.
[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]
Plugging in values:
[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]
b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)
- Tension force (upward, +)
The sum of these forces produces a centripetal force, upward (+).
[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]
Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]
Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]
A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 80.9 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
HINT
Answer:
Approximately [tex]4.61\times 10^{3}\; {\rm N}[/tex] upwards (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
External forces on this astronaut:
Weight (gravitational attraction) from the earth (downwards,) andNormal force from the floor (upwards.)Let [tex](\text{normal force})[/tex] denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:
[tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be [tex](\text{weight}) = m\, g[/tex].
Let [tex]a[/tex] denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be [tex](\text{net force}) = m\, a[/tex].
Rearrange [tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex] to obtain an expression for the magnitude of the normal force on this astronaut:
[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}[/tex].
The minimum takeoff speed for a certain airplane is 50 m/s. What minimum acceleration is required if the plane must leave a runway of length 2000 m? assume the plane starts from rest at one end of the runway.
Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that
[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]
Solve for [tex]a[/tex] :
[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]
Melanie gets out of her car at the park. She walks 25 m to the trail entrance.
She jogs around the trail until she reaches a pond, where she stops briefly.
She then continues to follow the trail around the pond. Which reference point
should be used to describe her motion?
A. Her car
B. The trail entrance
C. The pond
D. The trail
The reference point that should be used to describe Melanie motion is the pond (option C).
What is reference point?Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.
According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.
However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.
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calculate the pressure of water in a well if the depth of the water is 10m
Answer:
P= phg
so:- the height is 10m
density of water 1000 kg/m3
gravity is 9.8m/s2
P=1000 kg/m3*10m*9.8m/s2
=98000Pa
=98KPa
The form of energy that can move from place to place across the universe is
Answer:kinetic energy
Explanation:
You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
How much work do you perform during the lifting process?
8427 J is Incorrect.
If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?
80.25 W is Incorrect.
The total work done is 5980 Joules and the power expended is 57 Watts.
What is work done?
The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;
Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules
Height of the center of mass of chain = 25.9 / 2 = 12.95 m
Work done by the chain Wc;
Wc = 12.95 * 19.3 * 9.8 = 2450 Joules
Total work = 3530 + 2450 = 5980 Joules
Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts
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How do the Sun, Earth, and Moon systems interact
The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.
The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.
find the x-component of this vector.
The x-component for the vector given in the image is : [tex]B_{x}[/tex] = 18.3m
Meaning of a VectorA Vector can be defined as a quantity that possesses both magnitude and direction thereby giving more details.
A vector is very important when it comes to dealing with forces.
Analysis
[tex]B_{x}[/tex] = B cos(Θ)
Θ = 180° - 170° = 10°
[tex]B_{x}[/tex] = B cos(10°) = 18.3m
In conclusion, The x-component for this vector is : [tex]B_{x}[/tex] = 18.3m
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A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0°. What is the value of work being done on the object?
Answer:
The value of work being done on the object is 958J.
Explanation:
Work done is equal to force multiply by distance, but when the angle is between the force and distance work done=Force (cos theta) × distance
Work done = Force(cos theta) × distance
Work done = 25N(cos 40.0°) × 50m
Work done = 25N(0.7660) × 50m
Work done = 19.15N × 50m
Work done = 957.5J = 958J
Therefore the value of work being done on the object is 958J.
Q1) Write True or False for each statement:
Equal volumes of the two different substances have equal masses.
Answer:
True
Explanation:
it is true because blabla bla
9. What's the process that a spark represents?
A. A body with a strong positive charge is placed near a body with a neutral charge and the electrons jump between the two bodies
B. A body with a strong positive charge is placed near a body with a strong negative charge and the electrons jump between the two
C. A body with a neutral charge is placed near a body with a strong negative charge and the electrons jump between the two bodies
D. A body with a strong positive charge is placed near a body with a strong positive charge and the electrons jump hatupon the. The answer
Answer:
Sparks often indicate the presence of a high voltage, or "potential field". The higher the voltage; the farther a spark can jump across a gap, and with enough energy supplied can lead to greater discharges such as a glow or an arc.
Explanation:
in order for work to be done wht must happen?
Answer asap please due soon
An object must move in the same direction as the force is applied for work to be accomplished. Option C is correct.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative. it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The work done is found as;
W= F × d
Where.
F is the force
d is the displacement
W is the work done
In order for work to be done, an object must move in the same direction, and the force is applied
Hence option C is correct.
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A person of mass 100kg runs up a staircase with a vertical height of 10m. If the trips takes 5 seconds to complete, calculate the person's power ( acceleration due to gravity =10m/s²)
Answer:
Power = 2000J/s
Explanation:
Power= Total Work done/Time
Given
mass = 100kg
distance(height) = 10m
time = 5s
acceleration due to gravity = 10m/s²
Work done = force × distance
Finding Force
Force = mass × acceleration due to gravity
Force = 100kg × 10m/s²
Force = 1,000N
Finding work done
Work done = Force × distance
Work done = 1,000N × 10m
Work done = 10,000J
Finding Power
Power = work done/time
Power = 10,000J/5s
Power = 2,000J/s
Therefore Power is 2,000J/s
physical properties of matter help describe a substance which of the following is not a physical property of a soil sample?
A. the container in which the soul is held
B. the color of the soil
C. the volume of the soil
D. the texture of the soil
Answer asap this is an exam il give brainlist
An object moves in uniform circular motion at 50 m/s and takes 1.0 second to go a quarter circle. Calculate the
centripetal acceleration.
Answer:
The centripetal acceleration (ac)=314m/s²
Explanation:
look at the attachment ☝️
Kim has her coffee cup on her car's dash when she takes a corner with radius 4 m and 20 km/h.
What is the minimum coefficient of static friction which would allow the coffee cup to stay there
without slipping?
The minimum coefficient of static friction is 0.807
What is static friction?Static friction is friction between two or more solid objects that are not moving relative to each other.
According to the question,
When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:
Centripetal force and friction force.
The friction coefficient between the object and the path is independent of the object's mass.
Given,
The radius of a corner is, r=4.16m
The turning speed of the car is, v
[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]
= 5.72 m/s
To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.
So, Fc=Ff
[tex]\frac{mv^2}{r}[/tex]=μmg
μ=[tex]\frac{v^2}{rg}[/tex]
Here,
Fc represents the centripetal force.
Ff represents the friction force.
μ represents the coefficient of static friction.
g represents the gravitational acceleration whose value is 9.8m/s²
By Substituting the values in the above formula, we get:
μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex] ≈ 0.807
Therefore,
The minimum coefficient of static friction is 0.807
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A ball of mass m is dropped. What is the formula for the impulse exerted on the ball from the instant it is dropped to an arbitrary time τ later? Ignore air resistance.
The formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.
Impulse exerted on the ballThe impulse exerted on the ball at any given instant is calculated as follows;
J = ΔP
J = Pf - Pi
where;
J is impulse exerted on the ballpf is final momentumpi is the initial momentumThus, the formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.
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In the first seconds of flight, the Saturn V rocket achieved an altitude of m, and a velocity of m/s. The rocket weighed approximately kg. What was the average power produced by the rocket?
The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
Average power produced by the rocketThe average power produced by the rocket is calculated as follows;
P = FV
where;
P is the average powerF is the force exerted by the rocketV is the velocity of the rocketThus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
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●
Dean is skateboarding. He starts at a speed of 0.5 m/s and speeds up to 4.0 m/s
to go off a ramp over a time of 5.0 seconds. What is Dean's acceleration?
Answer:
[tex]\huge\boxed{\sf a = 0.7 \ m/s^2}[/tex]
Explanation:
Given Data:Initial velocity = [tex]V_i[/tex] = 0.5 m/s
Final velocity = [tex]V_f[/tex] = 4 m/s
Time = t = 5 s
Required:Acceleration = a = ?
Formula:[tex]\displaystyle a =\frac{V_f-V_i}{t}[/tex]
Solution:Put the givens in the formula.
[tex]\displaystyle a=\frac{4-0.5}{5} \\\\a = \frac{3.5}{5} \\\\a = 0.7 \ m/s^2\\\\\rule[225]{225}{2}[/tex]