Answer:
Explanation:
A wonderful (NOT) mixed unit problem
convert mph to m/s
v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)
v = 4.4704 m/s
KE = ½(0.006)4.4704²
KE = 0.05995342848 J
KE = 0.06 J
explain the different conditions that can result in hot and cold lahars, and explain how lahars change the earth's surface?
A crane is lifting a 500 lb piano with a velocity of 2fts''. Express the power of the crane in hp. (1 hp = 550ft-lbs)
Answer:
goalindia
goalindiaGiven:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kg
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 W
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s
goalindiaGiven:Mass of car, m = 500 lb = 500 × 0.4536 = 226.8 kgPower output of crane, P = 1.82 hp = 1.82 × 746 = 1357.72 WTo Find:The velocity of the car.Calculation:- The Power can be given by the formula:P = F × v⇒ v = P/F⇒ v = P/mg⇒ v = 1357.72/(226.8 × 9.8)⇒ v = 1357.72/2222.64⇒ v = 0.61 m/s- So, the velocity of the car will be 0.61❤
Two positive charges of 1 mC and 10 mC are separated by a distance of 10 m. Find the
direction and the magnitude of electrostatic force between the charges. |Describe the
direction in terms of the charges attract each other," or "the charges repel each other."
mili C=1043 C
Answer:
900N, charges repel
Explanation:
F = KQq/d²
K = 9 × 10^9Nm²/C²
Q= 10mC = 10 × 10^-3C
q = 1mc = 1 × 10^-3C
d = 10m
F = ?
Force = (9 × 10^9 × 10 × 10^-3 × 1 × 10^-3)/10²
= 9 × 10²
= 900N
It will be an electrostatic force of repulsion since like charges(two positive charges) repel
The magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.
What is Coulomb's law of forces?Coulomb's law states that whenever the two charged particles are separated by a particular distance then there will be a force of attraction or repulsion acts between the charge.
The formula of coulomb's force will be given by
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]K=9\times 10^9\ \frac{Nm^2}{C^2}[/tex]
[tex]q_1= 10\times 10^{-3}\ C[/tex]
[tex]q_2=1\times 10^{-3}\ C[/tex]
[tex]\rm Distance \ d=10 \ m[/tex]
Now the force will be calculated as
[tex]F=(9\times 10^9)\dfrac{(10\times 10^{-3})\times (1\times 10^{-3})}{10}[/tex]
[tex]F= 900\ N[/tex]
The charges are positive in nature so they will repel each other
Hence the magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.
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PLEASE HELP ASAP
I need help understanding this question, form my guess I’m thinking it’s C but I’m not to sure
Answer:
Explanation:
Without the diagram, we have to bracket our answers.
Total energy is constant, so we eliminate options A and B.
C is correct if position 1 is closer to the star than position 2
D is correct if position 2 is closer to the star than position 1
Would you die if you take the car keys out of the ignition and throw them in the back seat while you coast down a mountain-side in an attempt to save fuel and look cool for the girl in said back seat.
Answer:
bro what.
Explanation:
... is this an actual physics question?
A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s
The speed of the cannon after firing is 1.2 m/s
This can be solved using the law of conservation of momentum.
From the law of conservation of momentum,
⇒ The momentum of the projectile is equal to the momentum of the cannon.
MV = P................ Equation 1⇒ Where :
M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.⇒ make V the subject of the equation
V = P/M.................. Equation 2From the question,
⇒ Given:
P = 600 kgm/sM = 500 kg⇒ Substitute these values into equation 2
V = 600/500V = 1.2 m/sHence, The speed of the cannon after firing is 1.2 m/s
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As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement of the box is a result of
Answer:
inertia
Explanation:
the mass of the box will want to continue on its straight line motion.
As centripetal force on the wheel treads causes the truck to turn, frictional force between the box and truck bed is insufficient to keep the box turning at the same rate that the truck turns, so relative velocity occurs between box and bed until the centripetal force is increased by contact with the truck side wall farthest from the center of curvature.
when a student comned his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened
His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.
Answer:The comb and hair have become charged with opposite charges.
Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.
An accessory such as a coupling that performs a mechanical rather than an electrical function is called?
Blonde hair is produced by inheriting double recessive alleles (bb) from one’s parents. What is the probability of their offspring having blonde hair?
plz help
Answer:
over 50% power ability
Explanation:
I believe it's half and half chance that The Offspring or the child will have that trait
A constant force of 12 N in the positive x direction acts on a 4.0-kg object as it moves from the origin to the point 6i-8j m. How much work is done by the given force during this displacement
Answer:
Explanation:
Only force in the direction of motion does work
The force acts in the x direction
The x change of position was 6 m
W = Fd = 12(6) = 72 J
The work done by the given force during this displacement is determined as 120 J.
Work done on the objectThe work done on the object is calculated as follows;
W = Fd
where;
d is the displacement of the objectd = √(6² + 8²)
d = 10 m
W = 12 x 10
W = 120 J
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SOMEONE PLEASE HELP MEEEEEEE
Answer:
The movement of Tectonic plates.
Explanation:
Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton
hits the ball back at a velocity of -44 m/s. Determine the impulse Josh
Hamilton's bat delivered to the baseball and enter to the nearest kg-m/s with
the correct sign.
The Impulse delivered to the baseball is 89 kgm/s.
To solve the problem above, we use the formula of impulse.
⇒ Formula:
I = m(v-u)................. Equation 1Where:
I = Impulse delivered to the baseballm = mass of the baseballv = Final velocity of the baseballu = initial speed of the baseballFrom the question,
⇒ Given:
m = 0.8 kgu = 67 m/sv = -44 m/s⇒ Substitute these values into equation 1
I = 0.8(-44-67)I = 0.8(-111)I = -88.8I ≈ -89 kgm/sNote: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.
Hence, The Impulse delivered to the baseball is 89 kgm/s.
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URGENTLY NEED HELP
This is a 2 part question
42) You pull downward with a force of 28 N on a rope that passes over a disk-shaped pulley of mass 1.2 kg and radius 0.075 m. The other end of the rope is attached to a 0.67-kg mass. (a) Is the tension in the rope the same on both sides of the pulley? If not, which side has the greater tension? (b) Find the tension in the rope on both sides of the pulley.
Answer:
The net force on the hanging mass is thus
2 T Mg 18 6.6 N 11.4 N
, enough to accelerate it upward at
17 m/s2
. The angular acceleration of the pulley is thus
2 2
Explanation:
The tension in the rope on both sides of the pulley 20.21N.
To solve this problem, we'll need to consider the forces acting on both sides of the pulley and apply Newton's second law of motion. Let's break it down step by step:
(a) Is the tension in the rope the same on both sides of the pulley?
No, the tension in the rope is not the same on both sides of the pulley. The side with the greater mass attached will experience a greater tension in the rope.
(b) Find the tension in the rope on both sides of the pulley:
Calculate the gravitational force on each mass:
Gravitational force on the pulley: Fpulley = mpulley * g, where mpulley is the mass of the pulley and g is the acceleration due to gravity (approximately 9.8 m/s²).
Gravitational force on the hanging mass: Fhanging = mhanging * g, where m_hanging is the mass of the hanging mass.
Calculate the net force on the pulley:
Net force on the pulley is the difference between the tension forces on either side: Fnet = TLeft - Tright.
Apply Newton's second law to the pulley:
For the pulley, Fnet = mpulley * a, where a is the acceleration of the pulley. Since the pulley is assumed to be massless, we can use the relationship a = α * r, where α is the angular acceleration and r is the radius of the pulley.
Use the relationship between linear acceleration and angular acceleration:
α = a / r.
Equate the torque due to the tension to the moment of inertia times the angular acceleration:
τ = I * α, where τ is the torque, I is the moment of inertia of the pulley, and α is the angular acceleration.
Substitute the expression for α and solve for the net tension:
Tnet = (τ / r) = (I * α) / r = (1/2 * mpulley * r² * α) / r = (1/2 * mpulley * r * a).
Now, substitute the expression for a from step 3 and solve for the net tension:
Tnet = (1/2 * mpulley * r * α) = (1/2 * mpulley * r * (a / r)) = (1/2 * mpulley * a).
Substitute the expression for a from step 2 (Fnet = mpulley * a) and solve for the net tension:
Tnet = (1/2 * Fnet).
Now, you can find the tensions on each side of the pulley:
Tleft = Tnet + Fhanging
Tright = Tnet - Fpulle
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If a rocket experiences an acceleration generated by the gravity force between the earth and itself, what is this acceleration if the rocket flies 1000 km above the ground and the Earth's radius is 6.378 * 10 ^ 6 * r m. We know the Earth has a mass of 5.97*10^ 24 kg(in m/s^ 2 , G=6.67*10^ -11 N(m/kg)^ 2 ) ?
a 8.97
b 7.32
c 9.81
d 5.5
e 11.45
This question involves the concepts of Gravitational Force and Weight force.
The value of acceleration is "b. 7.32 m/s²".
At the given height the weight of the rocket must be equal to the gravitational force between rocket and the Earth:
[tex]W=F_G\\mg=\frac{GmM}{R^2}\\\\g=\frac{GM}{R^2}[/tex]
where,
g = acceleration = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = mass of earth = 5.97 x 10²⁴ kg
R = Radius of Earth + Height = 6.378 x 10⁶ m + 1 x 10⁶ m = 7.378 x 10⁶ m
Therefore,
[tex]g=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{(7.378\ x\ 10^6\ m)^2}[/tex]
g = 7.32 m/s²
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A ball has the energy to move 30 m/s with the mass of 5. What is the energy
Someone help me please !!!! Will mark Brianliest !!!!!!!!!!!!!!!!
An object in free fall has a velocity of 5m/s in the upward direction. What is the instantaneous velocity of the object one second later?
Answer:
Explanation:
Near the earth's surface where gravity is approximately 10 m/s² downward
v = u + at
v = 5 + (-10)(1) = -5 m/s
so it has the same speed but in the opposite (downward) direction.
The instantaneous velocity of the object is -4.8 m/s and is directed downwards.
What is meant by instantaneous velocity ?Instantaneous velocity is defined as the rate of change of position for a given interval of time. And it also represents the direction.
Here,
The object is freely falling.
So, the acceleration of the object is the acceleration due to gravity.
a = g = 9.8 m/s²
Initial velocity of the object, u = 5 m/s (upward velocity)
Time given, t = 1
Using the first equation of motion,
v = u + at
v = 5 + (-9.8) x 1
( negative sign is due to it is directed downwards which is opposite to u)
v = -4.8 m/s
That means the instantaneous velocity is directed downwards.
Hence,
The instantaneous velocity of the object is -4.8 m/s and is directed downwards.
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.
A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s.
What is the average speed of the object?
Answer:
Explanation:distance-time =speed
a,V =s/t
V=2m/2s
V=1m/s
b,v=s/t
V=80m/40s
V= 2m/s
The average speed is 2m/s
A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder
Answer:
Explanation:
I will ASSUME that the cylinder floats with the axis through the center of its flat ends is perpendicular to the water surface.
The cylinder will displace it's mass in water.
water density has a value of 1000 kg/m³
if the depth of the bottom is "h" in meters
1.80 kg = π(0.018² m²)(h m)(1000) kg/m³
h = 1.768388256...
or about 1.77 m
True or False. Isaac at first resisted his father's command to be the sacrifice.
Write down the condition required for the thermonuclear fusion.
Answer:
The main condition necessary for a controlled thermonuclear fusion:
The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.
) Water flows through the system shown below. Estimate the volumetric flow rate if friction
losses are negligible.
The volumetric flow rate if friction is negligible = 0.00197 m³/sec
Given data :
P₁ = 7 atm = 709.275 kPa
P₂ = 1.5 atm = 151.988 kPa
d₁ = 0.74 cm = 0.0074 m
d₂ = 1.20 cm = 0.0120 m
h = 82 m
density of water ( p ) = 1000 kJ/m³
Estimate the volumetric flow rate
given that friction losses are negligible
V₁² - V₂² = [tex]\frac{2(P_{2} - P_{1}) }{p} + 2g(h )[/tex]
= 2( 151.988 - 709.275 ) + 2 * 9.81 * 82
= 2 ( - 557.287 ) + 1608.84
= - 1114.574 + 1608.84 = 494.266
∴ V₁² - V₂² = 494.266 ----- ( 1 )
Also given that:
volumetric flow rate ( Q ) = A₁V₁ = A₂V₂
= π / 4 ( 0.0074 )² * V₁ = π / 4 ( 0.0120 )² * V₂
= 0.000043 V₁ = 0.000113 V₂
∴ V₁ = 2.63 V₂
Back to equation ( 1 )
( 2.63 v₂ )² - V₂² = 494.266
V₂² = ( 494.266 ) / 1.63
∴ V₂ = √303.23 = 17.41 m/sec
Therefore the volumetric flow rate ( Q )
Q = A₂V₂
= π / 4 * ( 0.0120 )² * 17.41
= 0.00197 m³/sec
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17. The shape of a river is determined by which of the following? (Choose all that
apply)
A Erosion
B. Deposition
C. Weathering
D. water speed
Answer:
Water speed Plays a key role In determining The shape of it
The ability to do work and cause change is called
Answer:
I am pretty sure it is called energy.
23 A 2-kg rock is moving at a speed 6 m/s, What constant force is needed to stop the rock in 7x10-52
Answer:
Explanation:
Do you mean 7 x 10⁻⁵² seconds? Who can even measure that small a time period?
If that time is incorrect, just redo the math with the correct time to get the force.
An impulse will cause a change of momentum
FΔt = mΔv
F = mΔv/Δt
F = 2(0 - 6)) / (7 x 10⁻⁵² - 0)
F = - 1.7 x 10⁵² N
The magnitude of that force would pretty much crush any rock it hit.
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Answer:
Decant it.
Explanation:
Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.
Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.
What is the significance of Isaac's name? What is the significance of Isaac's name?
Answer:
Isaac is one of the three patriarchs of the Israelites and is an important figure in the Abrahamic religions, including Judaism, Christianity, and Islam. Isaac's name means "he will laugh", reflecting the laughter, in disbelief, of Abraham and Sarah, when told by God that they would have a child.Hope this helps !!Explanation:
Itz Deepika Here to help ~ .
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1. An airplane flies with an airspeed (speed relative to the air) of 215
km/h. The wind is blowing at 45.0 km/h. Find the velocity of the
plane (relative to the ground) if the plane is pointed straight west and
the wind is:
A. blowing toward the
west (3 points)
B. blowing toward the east (3 points)
C. blowing toward the south (4 points)
(For part C be sure to calculate the angle as well.)
Answer:
Explanation:
A) 215 + 45.0 = 260 km/h West
B) 215 - 45.0 = 170 km/h West
C) √(215² + 45.0²) = 219.6588... 220 km/h
θ = arctan(45.0/215) = 11.8214... 11.8° S of W
The airplane's ground velocity is westward at 170 km/h regardless of wind direction. The south wind does not affect the plane's westward velocity in component C.
Vector addition will determine each scenario's plane velocity relative to the ground. We'll simplify by assuming a consistent horizontal breeze.
Let's divide the airplane's velocity and wind velocity into westward (negative) and eastward (positive) components.
Data: Plane airspeed = 215 km/h.
45 km/h wind.
A. West wind:
The airplane's westward (negative) ground velocity:
West aeroplane velocity = -215 km/h
-45 km/h west wind.
Find Vg when the wind blows west:
Vg (west) = Aeroplane velocity + Wind velocity (west)
Vg(west) = -215 km/h - (-45 km/h)
Vg (west)=-170 km/h
B. East wind: 2. Westward (negative) aeroplane velocity relative to the ground:
West aeroplane velocity = -215 km/h
Wind velocity (east) = 45 km/h (because the wind is blowing eastward)
Find Vg when the wind blows east:
Vg (west) = Aeroplane velocity (west) + Wind velocity (east).
Vg (west) = -215 + 45 km/h
Vg (west)=-170 km/h
C. Southerly wind: 3. Westward (negative) aeroplane velocity:
West aeroplane velocity = -215 km/h
Wind velocity (south) = 0 km/h (no effect since wind is not moving westward)
Find Vg when the wind blows south:
Vg (west) = Aeroplane velocity (west) + Wind velocity (south).
Vg (west) = -215 + 0 km/h.
Vg (west)=-215 km/h
All resultant velocities are westward and 170 km/h.
Part C's plane's westward velocity is unaffected by the south wind. The plane's ground velocity remains -215 km/h (westward). Since the heading is westward, no angle is needed.
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You are pulling a sled with a constant velocity using a rope held horizontally in
the snow. Using a spring scale attached to the rope, you measure the tension
to be 0.96 N. The sled's mass is 0.58 kg. Write the surface force the ground
exerts on the sled in component form, and determine its magnitude.
Answer:
Explanation:
if you are pulling in the positive direction and Upward is also positive.
F = -0.96i + 0.58(9.8)j
F = -0.96i + 5.7j
F = √(-0.96² + 5.7²) = 5.8 N
You are pulling a sled with constant velocity by using a rope then the surface force the ground exerts on the sled will be 5.8 N.
What is tension?The pulling force conveyed axially by a string, cable, chain, or another analogous object, or by either end of a rod, truss member, or another such three-dimensional object, is referred to as tension in physics. Another way to think of tension is as the action-reaction pair of forces acting at each end of the previous elements. Perhaps tension is the polar opposite of compression.
A restoring force may cause what is also referred to as tension when atoms or molecules are torn apart from one another at the atomic level and build up potential energy.
If you are moving upward and pushing in a positive way,
F = -0.96 i + 0.58(9.8) j
F = -0.96 i + 5.7 j
F = [tex]\sqrt{(-0.96^2 + 5.7^2)}[/tex]
F= 5.8 N
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