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
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resistance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter
Answer:
Explanation:
For resistance , the expression is as follows .
R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .
cross sectional area = π x ( .5 x 10⁻³ )²
S = .785 x 10⁻⁶ m²
Putting the values
R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶
= 1.796 x 10⁻² ohm .
You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Object 3 is a sphere of copper.
a. the density of tin is 5.75g/cm2. What is the mass of object 1 in kg if the rectangular block has a volume of 1.34L?
b. what is the volume in cubic inches of object 2 if the cube of aluminum 7.58 inches on a side?
c. what is the mass in kg of object 2? the density of aluminum is 2.70g/cm3
d. what is the volume in cm3 of object 3 if the sphere of copper has a diameter 8.62cm? the volume of the sphere is 4 {pi}^3/3
e. what is the mass in kg of object 3? Copper has a density of 8.96g/cm3
Answer: a. m = 7.7 kg
b. V = 435.52 in³
c. m = 1927 kg
d. V = 335.37 cm³
e. m = 3 kg
Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.
The formula for density is
[tex]\rho=\frac{m}{V}[/tex]
Density's unit in SI is kg/m³, but it can assume lots of other units.
Some unit transformations necessary for the resolution of the question:
1 L = 1 dm³ = 1000 cm³
1 in³ = 16.3871 cm³
1 g = 0.001 kg
a. V = 1.34 L = 1340 cm³
[tex]\rho=\frac{m}{V}[/tex]
[tex]m=\rho.V[/tex]
m = 5.75 * 1340
m = 7705 g => 7.705 kg
Mass of object 1 with volume 1.34L is 7.7 kg.
b. A cube's volume is calculated as V = side³
V = 7.58³
V = 435.52 in³
Volume of object 2 is 435.52 in³.
c. Using 1 in³ = 16.3871 cm³ to change units:
V = 435.52 * 16.3871
V = 713689.4 cm³
Then, mass will be
[tex]m=\rho.V[/tex]
m = 2.7 * 713689.4
m = 1926961.4 g => 1927 kg
Mass of object 2 is 1927 kg.
d. Volume of a sphere is calculated as [tex]V=\frac{4}{3}.\pi.r^{3}[/tex]
Diameter is twice the radius, then r = 4.31 cm.
Volume is
[tex]V=\frac{4}{3}.\pi.(4.31)^{3}[/tex]
V = 335.37 cm³
Volume of object 3 is 335.37 cm³.
e. [tex]m=\rho.V[/tex]
m = 8.96 * 335.37
m = 3004.91 g => 3 kg
Mass of object 3 is 3 kg.
when air mass is caught between two cold fronts the result is a _______ front.
Answer choices
A.occluded
B.warm
C.cold
D.stationary
HELP ASAP PLEASE
Which 2 factors must be present for chemical vapor deposition to be successful:
A.) The size of the diamond is larger than the most.
B.) The conditions during cooling are controlled.
C.) The heat in a vacuum forms a gas of single atoms.
D.) The heat in a vacuum is decreased to freezing.
E.) The pressure of a reaction vessel is negative.
kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
0.2 mph
4.8 mph
5.5 mph
144.1 mph
it is actually science i couldn't find the word science
Answer:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
Answer:
38.7 mph
Explanation:
I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
What happens to the force attraction of the distance two objects is increased?
Answer:
Explanation:
The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.
what part of the electromagnetic spectrum can our skin detect?
Visible and infrared light.
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M and positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.
Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?
Answer:
a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Explanation:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by
E = s/2ε₀[1 - z/√(z² + R²)]
So, the net force on the small plastic sphere of mass M and charge Q is
F = QE
F = Qs/2ε₀[1 - z/√(z² + R²)]
b. At what height h does the sphere hover?
The sphere hovers at height z = h when the electric force equals the weight of the sphere.
So, F = mg
Qs/2ε₀[1 - z/√(z² + R²)] = mg
when z = h, we have
Qs/2ε₀[1 - h/√(h² + R²)] = mg
[1 - h/√(h² + R²)] = 2mgε₀/Qs
h/√(h² + R²) = 1 - 2mgε₀/Qs
squaring both sides, we have
[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²
h²/(h² + R²) = (1 - 2mgε₀/Qs)²
cross-multiplying, we have
h² = (1 - 2mgε₀/Qs)²(h² + R²)
expanding the bracket, we have
h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²
collecting like terms, we have
h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²
Factorizing, we have
[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²
So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]
taking square-root of both sides, we have
√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]
h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
what is a asteroid traveling rapidly called
Answer:
meteor
Explanation:
A asteroid stays still and a meteor goes fast
Answer:
meteor
Explanation:
or some people call it a shooting star
Lolliguncula brevis squid use a form of jet propulsion to swim—they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15m/s0.15m/s or greater, they can accelerate at 1.2m/s21.2m/s 2 .
(a) Determine the time interval needed for a squid to increase its speed from 0.15m/s0.15m/s to 0.45m/s0.45m/s.
(b) What other questions can you answer using the data?
Answer:
a) t = 0.25 s, b) x = 0.075 m
Explanation:
a) For this exercise we will use kinematic relationships in one dimension
v = v₀ + a t
in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²
t = [tex]\frac{v -v_o}{a}[/tex]
we calculate
t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]
t = 0.25 s
b) We can also find the distance traveled during this acceleration
v² = v₀² + 2a x
x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]
let's calculate
x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]
x = 0.075 m
Describe Kinetic Energy and Potential Energy (in your own words please!!)
Answer:
Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.
Explanation:
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
A bartender slides a beer mug at 1.3 m/s towards a customer at the end of a frictionless bar that is 1.3 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor
Answer:
x = 0.67 m
Explanation:
For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.
Let's find the time or it takes to get to the floor
y = y₀ + v_{oy} - ½ g t²
in this case I go = 0 and when I get to the floor y = 0
0 = y₀ + 0 - ½ g t²
t² = 2y₀ / g
t² = 2 1.3 / 9.8 = 0.2653
t = 0.515 s
now let's find the distance traveled in this time
x = vx t
x = 1.3 0.515
x = 0.6696 m
x = 0.67 m
why would the bulb not light?
are you a dmbss? the bulb and wire must be connected to both end
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
Two masses are being pulled up a 30.0-degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s2 and the velocity is down the incline. The force is applied to a 200-kg mass and a string connects the 200-kg mass to a 150-kg mass. The coefficient of kinetic friction is 0.200. The force F is
Answer:
Explanation:
The acceleration up the incline is 1.00 m/s²
Net force acting on two masses = total mass x acceleration
= 350 x 1 = 350 N
weight acting down the plane = m g sinФ
= 350 x 9.8 x sin30 = 1715 N
Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction
= 350 x 9.8 x cos30 x .2 = 594N
Net force acting on total mass
= F - 1715 - 594 = 350 , where F is required force
F = 2659 N .
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
Anyone know this question?
Where the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place _____ the lock installed.
Answer:
With or without.
Explanation:
According to the National Electrical Code (NEC), here the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place with or without the lock installed. Thus, this is in accordance with section 110.25 of the National Electrical Code (NEC).
What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
For each picture below identify the material and closing the sandwich as transparent translucent or opaque
In the absence of air resistance, ___ accelerates all objects at the same rate of 9.8 m/s2
Answer:
Gravity...this was proven by NASA and was examined by Leonardo da Vinci.
Answer:
Gravity
Explanation:
In the absence of air resistance, Gravity accelerates all objects at the same rate of 9.8 m/s2
Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2300 kg/s steadily. Determine the power generation potential.
Answer:
the power generation potential is 2.705 x 10⁶ J/s.
Explanation:
Given;
height below the free surface of a large water reservoir, h = 120 m
mass flow rate of the water, m' = 2300 kg/s
The power generation potential is calculated as;
[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]
Therefore, the power generation potential is 2.705 x 10⁶ J/s.
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
an always be used to calculate the electric field. relates the electric field at points on a closed surface to the net charge enclosed by that surface. relates the surface charge density to the electric field. relates the electric field throughout space to the charges distributed through that space. only applies to point charges.
Complete Question:
Gauss's law:
Group of answer choices
A. can always be used to calculate the electric field.
B. relates the electric field throughout space to the charges distributed through that space.
C. only applies to point charges.
D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.
E. relates the surface charge density to the electric field.
Answer:
D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.
Explanation:
Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.
This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.
This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.
Mathematically, Gauss's law is given by this formula;
ϕ = (Q/ϵ0)
Where;
ϕ is the electric flux.
Q represents the total charge in an enclosed surface.
ε0 is the electric constant.