Answer:
Kinetic
Explanation:
Because it has a force pushing on it
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign
Explanation:
According to newton's second law of motion.
[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]
m is the mas of the sky diver = 93.4kg
a is the acceleration of the skydiver
From the formula above;
[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]
Hence the acceleration of the sky diver is 1.563m/s²
Two equal forces act on two different objects, one of which has a mass ten times as large as the other. The larger object will have _________ acceleration that the less massive object.
Answer:
The larger object will have smaller acceleration that the less massive object.
Explanation:
Generally force is mathematically represented as
[tex]F = ma[/tex]
=> [tex]m = \frac{F}{a }[/tex]
at constant force we have
[tex]m \ \alpha \ \frac{1}{a}[/tex]
So if m is increasing a will be decreasing which means the object with the larger mass will have less acceleration
. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.
Answer:
Explanation:
The total distance is how far you walk from the starting point.
Distance through west = 18.0m
Distance through north = 25.0m
Total distance covered = 18.0+25.0m
Total distance covered = 43.0m
This means that I am 43.0m from the starting point
Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:
[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]
The direction of your displacement is 30.81m
Direction is gotten according to the formula;
[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]
Note that the direction to the west is negative, that is why the x is -18.0m
The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
Given-
Distance travel through the west is 18 m.
Distance travel through the north is 25 m.
Distance from starting point-
To know the total distance, add both the covered distance. Thus total distance x is,
[tex]x=18+25[/tex]
[tex]x=43[/tex]
Hence, the distance from the starting point is 43 m.
The displacement vector-
Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,
[tex]d^2=18^2+25^2[/tex]
[tex]d^2=324+625[/tex]
[tex]d^2=949[/tex]
[tex]d=\sqrt{949}[/tex]
[tex]d=30.81[/tex]
The displacement vector is 30.81 m.
The Direction of displacement-The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,
[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]
Here due to the west direction(opposite side), the sign is taken negatively.
[tex]\theta=tan^{-1}(-1.389)[/tex]
[tex]\theta=-60.27^o[/tex]
For the other quarter,
[tex]\theta=180-60.27=119.7^o[/tex]
Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
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A coin rests on a record 0.15 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient of static friction between the coin and the record is 0.30.
Required:
What is the maximum coin speed at which it does not slip?
Answer:
0.66m/sExplanation:
We are expected to solve for the velocity with no slip condition
we know that the expression that relate coefficient of friction and velocity is given as
μs = v^2/rg
Given
coefficient of friction μs = 0.3
radius r= 0.15
assume g=9.81m/s^2
substituting into the expression we have
0.3= v^2/0.15*9.81
v^2=0.3*0.15*9.81
v^2=0.44145
v=√0.44145
v=0.66
therefore the velocity is 0.66m/s
A student rides a bicycle 2400 meters in four minutes to get to school. What is the student's speed?
Answer:
10 mls2
Explanation:
speed =distance /time
Water enters a 2 m3 tank at a rate of 6 kg/s and is withdrawn at a rate of 2 kg/s. The tank is initially half full. What type of process is this?
Answer:
This process is semicontinuous.
Explanation:
Given that,
Volume = 2 m³
Enter flow rate = 6 kg/s
Exit flow rate = 2 kg/s
The tank is initially half full.
We need to find what type of process
Using given data,
This process is not continuous because given enter and exit flow rate is not equal.
This process is semicontinuous and the water level in the tank does not reach a constant level.
Hence, This process is semicontinuous.
why are elements important
Answer: Scientists believe that about 25 of the known elements are essential to life. Just four of these – carbon, oxygen, hydrogen, and nitrogen– make up about 96% of the human body. Six common elements that are important in living things are carbon, hydrogen, oxygen, nitrogen, sulfur, and phosphorus. These large molecules make up the structures of cells and carry out many many processes essential to life.
Plz mark brainliest:)
An 10.2-kg stone at the end of a steel (Young's modulus 2.0 x 10^11 N/m2) wire is being whirled in a circle at a constant tangential
speed of 11.6 m/s. The stone is moving on the surface of a frictionless horizontal table. The wire is 3.62 m long and has a radius of
4.10 x 10^-3m. Find the strain in the wire.
Answer:
[tex]3.6\times 10^{-5}[/tex]
Explanation:
m = Mass of stone = 10.2 kg
v = Tangential velocity = 11.6 m/s
l = Length of wire = 3.62 m
r = Radius of wire = [tex]4.1\times 10^{-3}\ \text{m}[/tex]
A = Area of wire = [tex]\pi r^2[/tex]
Y = Young's modulus of steel = [tex]2\times 10^{11}\ \text{N/m}^2[/tex]
[tex]\varepsillon[/tex] = Strain
The force acting on the stone will be centripetal
[tex]F=\dfrac{mv^2}{l}\\\Rightarrow F=\dfrac{10.2\times 11.6^2}{3.62}\\\Rightarrow F=379.15\ \text{N}[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{379.15}{\pi (4.1\times 10^{-3})^2}\\\Rightarrow \sigma=7179488\ \text{N/m}^2[/tex]
Young's modulus is given by
[tex]Y=\dfrac{\sigma}{\varepsilon}\\\Rightarrow \varepsilon=\dfrac{\sigma}{Y}\\\Rightarrow \varepsilon=\dfrac{7179488}{2\times 10^{11}}\\\Rightarrow \varepsilon=3.6\times 10^{-5}[/tex]
Strain in the wire is [tex]3.6\times 10^{-5}[/tex].
Describe the path of Earth if the Sun's gravity were to suddenly stop.
Answer:
The Earth is traveling around the sun with an orbital velocity of 30 kilometers per second. This is exactly the speed it needs to be going to counteract the force of gravity from the sun pulling it inward. If the sun were to suddenly disappear, Earth would travel in a perfectly straight line at 30 km/s
Explanation:
Magnetic attraction is one of the chemical properties of matter *
True
False
How will the motion of an object that is moving to the right change, if it is pushed in the opposite direction with a greater force?
Оооо
The object will move at a constant speed in the same direction for a while and then slow down and stop.
The object will slow down for a while and then move at a slower constant speed in the same direction.
The object will slow down and then begin to move faster and faster in the opposite direction.
0 The object will speed up and then begin to move at a slower speed in the opposite direction.
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
plzzz help its for my civics class what answer is it????????????
Answer:
the correct answer is The English Bill of Rights
Explanation:
This is the answer because i have taken the test.
A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)
Answer:
Explanation:
The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .
a )
Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .
b )
magnetic field created at the magnetic needle B = 10⁻⁷ x 2I / d where I is current and d is distance .
B = 10⁻⁷ x 2 x 26.3 / .27
= 194.81 x 10⁻⁷ T
angle of deflection of solenoid = 22.9°
Tan 22.9 = B /H
.422 = 194.81 x 10⁻⁷ / H
H = 461.63 x 10⁻⁷ T
= .46 x 10⁻⁴ T .
A) The current in the wire flows towards the Earth's surface
B) The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface
B) Determine The magnitude of the horizontal component of the Earth's magnetic field
B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )
where : l = 26.3 A, d = 0.27 m
Back to equation ( 1 )
B = 10⁻⁷ * 2 * 26.3 / 0.27
= 194.81 * 10⁻⁷ T
Final step : Calculate the magnitude of horizontal component ( H )
Tan ∅ = B / H ---- ( 2 )
where : ∅ ( angle of deflection ) = 22.9°
∴ H = B / Tan ( 22.9° )
= ( 194.81 * 10⁻⁷ ) / 0.422
= 0.46 x 10⁻⁴ T
Hence we can conclude that The current in the wire flows towards the Earth's surface and The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
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A sharpshooter fires a 0.22 caliber rifle horizontally at 100 m/s at a target 75m away. How far does the
bullet drop by the time it reaches the target?
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The bullet drops "2.76 m" by the time it reaches the target.
First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:
[tex]s = vt\\\\t = \frac{s}{v}[/tex]
where,
s = distance = 75 m
v = velocity = 100 m/s
t = time = ?
Therefore,
[tex]t = \frac{75\ m}{100\ m/s}[/tex]
t = 0.75 s
Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.
[tex]h = v_it+\frac{1}{2}gt^2[/tex]
where,
h = height dropped = ?
vi = initial vertical speed = 0 m/s
t = time interval = 0.75 s
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2[/tex]
h = 2.76 m
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The attached picture shows the equations of motion in the horizontal and vertical directions.
A force of 40 N is applied tangentially to the rim of a solid disk of radius 0.10 m. The disk rotates about an axis through its center and perpendicular to its face with a constant angular acceleration of 145 rad/s2. Determine the mass of the disk.
Answer:
m = 2,759 kg
Explanation:
For this exercise we use the torque relationship
τ = I α
the moment is
τ= F r sin θ
since the force is tangential to the ring, the angle is 90º sin 90 = 1
τ = F r
the moment of inertia of a ring is given by
I = m r²
let's substitute
F r = m r²α
m = F / r α
let's calculate
m = 40 / (0.10 145)
m = 2,759 kg
g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at the other. The cylinder has a radius of 0.0200 m and the safety valve is simply a 0.00750-m radius circular opening at one end, sealed with a disk. The disk is held in place by a spring with a spring constant of 950 N/m that has been compressed 0.0085 m from its natural length. Determine the magnitude of the minimum force that must be exerted on the piston in order to open the safety valve.
Answer:
58.32 N
Explanation:
Area of a circle = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the circle.
The cylinder has a radius of 0.02 m, its area is;
[tex]A_{1}[/tex] = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](0.02)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x 0.0004
= 1.2571 x [tex]10^{-3}[/tex]
Area of the cylinder is 0.0013 [tex]m^{2}[/tex].
The safety valve has a radius of 0.0075 m, its area is;
[tex]A_{2}[/tex] = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](0.0075)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x 5.625 x [tex]10^{-5}[/tex]
= 1.7679 x [tex]10^{-4}[/tex]
Area of the valve is 0.00018 [tex]m^{2}[/tex].
From Hooke's law, the force on the safety valve can be determined by;
F = ke
[tex]F_{2}[/tex] = 950 x 0.0085
= 8.075 N
Minimum force, [tex]F_{1}[/tex], required can be determined by;
[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
[tex]\frac{F_{1} }{0.0013}[/tex] = [tex]\frac{8.075}{0.00018}[/tex]
[tex]F_{1}[/tex] = [tex]\frac{0.0013 *8.075}{0.00018}[/tex]
= 58.32
The minimum force that must be exerted on the piston is 58.32 N.
the peripheral nervous system is responsible for both sending and receiving signals to and from the brain
Answer:
its true trust me
Explanation:
Answer: true
Explanation: edge
If a car is traveling at an average speed of 70 kilometers per hour how long does it take the car to travel 14 kilometers
Answer:
Explanation:
O.20 hour A
If a car is traveling at an average speed of 70 kilometers per hour, 0.2 hours it takes the car to travel 14 kilometers.
What is average speed?By multiplying the distance that an item travels in one unit by the amount of time it takes to go that distance, one may determine the speed of the object. The speed of the item on this voyage, denoted by the letter "s," is equal to s = D/T if "D" is indeed the distance traveled in certain time "T."
Understanding average speed will help you better comprehend the pace of a travel. On a travel, the pace could occasionally change. Knowing the average speed then becomes crucial to getting an idea of how quickly the route will be finished.
Distance covered = average speed × Time travelled
14=70× Time travelled
Time travelled = 0.2 hours
Therefore, 0.2 hours it takes the car to travel 14 kilometers.
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4. A teacher throws the stapler at her students 23 m in 1.2 seconds. What speed must the teacher throw the
stapler and what will be the staplers highest point during its flight?
dy = - 42gt2
dx = Vyt
Answer:
dy
Explanation:
My guess of guts.
okay bye
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
help me get the answer in Physical Science.
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
A Navy Seal of mass 80 kg parachuted directly down into an enemy harbor. At one point while she was falling, the resistive force that air exerted on her was 520 N upward. What can you determine about her motion at this point in time
Answer:
The Navy Seal is accelerating downwards at the rate of 3.3 m/s²
Explanation:
Given;
mass of the Navy Seal, m = 80 kg
the upward resistive force on her, F = 520 N
Her net downward force is given by;
[tex]F_{net} = F_{down} - F_{up}\\\\F_{net} = (80*9.8) - 520\\\\F_{net} = 264 \ N[/tex]
Her downward acceleration at this time is given by;
F = ma
a = F / m
a = 264 / 80
a = 3.3 m/s²
Therefore, the Navy Seal is accelerating downwards at the rate of 3.3 m/s²
a jogger travels at 4 m/s for 100 s what is the distance covered
400m
Explanation:
given,
v= 4m/s
t= 100s
d= ?
since, v = d / t
therefore, d = v * t (velocity multiplied by time)
=> d = 4 * 100
= 400m.
Which kind of energy do flowing electrons have?
A. Electrical
B Elastic
C. Nuclear
D. Gravitational
A lamp of mass m hangs from a spring scale which is attached to the ceiling of an elevator. When the elevator is stopped at the fortieth floor, the scale reads mg. What does it read as the elevator slows down to stop at the ground floor?
a. more than mg
b. mg
c. less than mg
d. zero
e. can't tell
Answer:
The correct answer is (a)
Explanation:
A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.
If F = ma, as a increases, F will also increase and vice versa
Where F = force
m = mass
a = acceleration (due to gravity in this case)
From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.
The emf of the battery is 1.5 V. In Nichrome there are 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is 7 × 10−5 (m/s)/(N/C). Each thick wire has length 29 cm = 0.29 m and cross-sectional area 9 × 10−8 m2. The thin wire has length 6 cm = 0.06 m and cross-sectional area 1.3 × 10−8 m2. (The total length of the three wires is 64 cm.) In the steady state, calculate the number of electrons entering the thin wire every second. Do not make any approximations, and do not use Ohm's law or series-resistance equations.
Answer:
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Explanation:
Given;
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
The magnitude of the electric field in the thin wire is given by;
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
the number of electrons entering the thin wire every second is given by;
[tex]e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s[/tex]
Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Calculation of the number of electrons:Since
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
So here the magnitude should be
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
Now the number of electrons should be
= 7 × 10⁻⁵ *25
= 1.75 x 10⁻³ mobile
hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
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A projectile is fired straight up with an initial velocity of 40.0 m/s . Approximately how high will the projectile ?
Answer:
it depends on the wind and any other conditions but if you have a controlled environment it should take 1 second to get 40 meters but it could go higher in which it could take about 5 seconds to go 200 meters
Explanation:
hope it helped
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Momentum of the 2 kg mass moving with velocity 10 m/s is *
A. 2 kg*m/s
B. 20 kg*m/s
C. 200 kg*m/s
D. 20000 kg*m/s
The forces exerted on an object are shown. (3 points)
A box has an arrow pointing up labeled F and an arrow pointing down labeled 3 N.
If the net force on the object along the vertical plane is zero, which statement is correct?
F equals 3 N and the object moves up.
F equals 3 N and the object remains stationary.
F equals 0 N and the object moves down.
F equals 0 N and the object remains stationary.
Answer:
F equals 3 N and the object remains stationary. (second option in the list)
Explanation:
For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.
Answer: F equals 3 N and the object remains stationary.
Explanation: