Answer:
11.573
Explanation:
f = m*a
where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration
so, we solve for a
a = f/m
a = 22/1.901
a = 11.573
Which of these is a push or a pull? Acceleration Force Mass Inertia
Answer:
the answer is force . force is applied as a push or pull
Word Bank:
Electrical
Mechanical
Chemical
Light
Thermal
Sound
Answer:
Light to
Electrical to
mechanical and sound
If you have to measure the temperature > above 80°c, do you use an alcohol or mercury thermometer? why?
Answer:
I use mercury thermometer to measure the temperature above 80°C because the boiling point of alcohol is 78°C . So, it can't measure the temperature above 78°C.
2. A ray of light is incident at 60° in the air on an air glass plane surface find the angle of refraction in the glass. (mew for glass=1.5)
[tex] [/tex]
Answer:
35.2644
I suppose mew is refractive index
Explanation:
( sin i ) / (sin r) = refractive index
( sin 60) / (sin r) = 1.5
( sin 60) / 1.5 =sin r
r=35.844
sorry if I'm wrong
A volcano launches a lava bomb straight upward with an initial speed of 24 m/s. speed at 2 and 3 seconds and it it is upward or downward
Answer:
Explanation:
v = u + at
Let Up be the positive direction
v(2) = 24 + (-9.8)(2) = 4.4 m/s Positive result means Upward
v(3) = 24 + (-9.8)(3) = -5.4 m/s Negative result means Downward
Which of the following is an incorrect statement?
The pulley is a special type of wheel and axle.
The mechanical advantage of a pulley with three ropes is one.
The mechanical advantage of a block and tackle pulley with two pulleys is two.
The mechanical advantage of a fixed pulley is one.
PLZ HELP WILL MARK BRAINLIEST
Answer:
Explanation:
The mechanical advantage of a pulley with three ropes is one.
is incorrect. It is 3
A roll of kitchen aluminum foil is 30 cm wide by 22 m long (if you unroll it). If the foil is 0.15 mm thick, and the specific weight of aluminum is 26460 N/m3, how much does the roll of aluminum foil weigh
The weight of the aluminum foil is 26.20 N
To find the weight of the aluminum foil when it is unrolled, we need to find its volume.
The volume V = lwt where
l = length of aluminum foil = 22 m, w = width of aluminum foil = 30 cm = 0.30 m and t = thickness of aluminum foil = 0.15 mm = 0.15 × 10⁻³ m.So, V = lwt
= 22 m × 0.30 m × 0.15 × 10⁻³ m
= 0.99 × 10⁻³ m³.
So, its weight W = ρV where
ρ = specific weight of aluminum = 26460 N/m³ and V = volume of aluminum foil = 0.99 × 10⁻³ m³So, W = ρV
W = 26460 N/m³ × 0.99 × 10⁻³ m³
W = 26195.4 × 10⁻³ N
W = 26.1954 N
W ≅ 26.20 N
So, the weight of the aluminum foil is 26.20 N
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What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25
s with 7,500 N of force?
128 kg
3991 kg
0.017 kg
14,091 kg
The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.
HOW TO CALCULATE MASS:
The mass of an object can be calculated by dividing the force applied to the object by its acceleration. According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:a = (v - u)/ta = 15.5 - 0/8.25a = 15.5/8.25a = 1.88m/s²The mass of the bus = 7500N ÷ 1.88m/s²The mass of the bus = 3989.4kgTherefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.Learn more about mass at: https://brainly.com/question/20259048?referrer=searchResults
AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.
A ball is launched from ground level at 30 m/s at an angle of 35° above the horizontal. how far does it go before it is at ground level again
Answer:
Explanation:
Ignoring air resistance
Initial vertical velocity is 30sin35 = 17.2 m/s
Gravity reduces this velocity to zero in a time of
t = v/g =17.2 / 9.8 = 1.755 s
it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s
The horizontal velocity is 30cos35 = 24.57 m/s
the distance traveled horizontally is
d = vt = 24.57(3.51) = 86.298... = 86 m
n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun
Answer:
F = M2 ω^2 R centripetal force of sun on planet
ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period
P = 2 pi (M2 * R / F)^1/2
F = G M1 M2 / R^2 gravitational force on planet
P = 2 pi {R^3 / (G M1)]^1/2
P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2
P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec
1 yr = 3600 * 24 * 365 = 3.15E7 sec
P = 3.9 / 3.2 = 1.2 years
A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m when it begins to roll down the incline. If rolling and sliding friction are neglected, what is the linear velocity, in m/s, of the center-of-mass of the wheel when it reached the bottom of the incline?
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
The motor of an electric drill has a power input of 1200 W. How much work 3 points
is done by the drill in a time of 2 minutes? *
Your answer
This is a required question
Answer:
Explanation:
1 Watt = 1 J/s
1200 J/s(2 min)(60 s/min) = 144 KJ
Two a.c V1=100sin(wt) and V2 = 150cos(wt) are fed into one circuit, determine the combined output of the two as a single a.c
Answer:
Explanation:
V = 100sin(ωt) + 150cos(ωt)
let x = ωt
V = 100sin(x) + 150cos(x)
a maximum or minimum will occur when the derivative is zero
V' = 100cos(x) - 150sin(x)
0 = 100cos(x) - 150sin(x)
100cos(x) = 150sin(x)
100/150 = sin(x)/cos(x)
0.6667 = tan(x)
x = 0.588 rad
V = 100sin(0.588) + 150cos(0.588)
V = 180.27756
as the maximum will not occur until ωt = 0.588 radians, for a cosine function we subtract that amount as a phase angle φ
V = 180.3 cos(ωt - 0.588)
or as a sine function, the phase angle lags the cosine by a difference of π/2
V = 180.3sin(ωt - (0.588 - π/2)
V = 180.3sin(ωt + 0.983)
please help
Two masses m1 and m2, are a distance R apart and mq exerts a gravitational force F on m2. What is the gravitational force is the gravitational force on M1?
Answer:
F
Explanation:
for every action there exists an equal and opposite reaction. The force acting on the two masses is identical in magnitude, but opposite in direction.
2) A traffic light of weight 100 N is supported by two ropes as shown. Let T1 and T2 are the tensions.
a. Resolve the vectors into its components and find their values [ 4 marks]
b. Find sum of the x-components [1 mark]
c. Find sum of the y-components [1mark]
d. Use the above equations to find the tensions in the ropes?
the two angles on x and y are both 37 I cant upload the diagram
Hi there!
a.
We know that:
[tex]\Sigma F_y = 0 \\\\\Sigma F_x = 0[/tex]
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W
[tex]T_1 = T_2 = \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}[/tex]
A gazelle is grazing while standing in a fixed location. When it's startled by a predator, the gazelle accelerates uniformly for 16.8 s until it reaches a speed of 21.9 m/s. The gazelle then runs in a straight line at constant speed for an additional 16.4 s. Finally, it uniformly slows to a stop at a rate of 1.8 m/s/s. What is the total distance traveled by the gazelle in meters
Answer:
Explanation:
acceleration phase
average speed was 21.9/2 = 10.95 m/s
distance covered is 10.95 m/s(16.8 s) = 183.96 m
distance at top speed 21.9 m/s(16.4 s) = 359.16 m
distance while decelerating (0² - 21.9²)/(2(-1.8)) = 133.225 m
total = 183.96 + 359.16 + 133.225 = 676.345 = 676 m
What does the horizontal line through the center of the wave on a graph represent?
Answer:
This is the midline or the medium which is the exact middle of the graphs minimum and maximum points(which are the amplitude)
How much tension must a rope withstand if it is used to accelerate a 1710-kgkg car horizontally along a frictionless surface at 1.30 m/s 2 m/s2
Answer:
Explanation:
Lowest tension will occur when the rope is also horizontal.
F = ma
F = 1710(1.30)
F = 2223 N
the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n
The ratio of the primary turns to the secondary turns is 1/3
The correct answer to the question is Option A. 1/3
From the question given above, the following data were obtained:
Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?Ratio = Nₚ/Nᵣ
Nₚ/Nᵣ = 225 / 675
Nₚ/Nᵣ = 1/3
Therefore, the ratio of the primary turns to the secondary turns is 1/3
Complete question:
See attached photo
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A 10kg object is 15 meters up a hill. Find its potential energy
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
A section of a rollercoaster is shown below. If the 17,500 kg rollercoaster momentarily comes to rest at point A, where it is 19 m in the air, how fast is it going at point D?
Answer:
Explanation:
Impossible to say without knowing how high point D is.
If we ignore friction, the energy converted from potential energy will exist as kinetic energy.
let d be the height in meters of point D "in the air"
mg(19 - d) = ½mv²
m is common so divides out
g(19 - d) = ½v²
v = √(2g(19 - d))
no sense in making g any more precise than the height. g = 9.8
v = √(2(9.8)(19 - d))
v = √(19.6(19 - d))
In Newton's cannonball experiment, if the velocity is equal the orbital velocity the Cannon ball
will
O stay in orbit
O fall to the earth
O escape the earth
O none of the above
In Newton's cannonball experiment, if the velocity is equal to the orbital velocity then the cannonball will stay in Orbit.
Newtons cannonball experiment stated that the distance that a cannonball will travel, before being drawn into the Earth by the forces of gravity, is dependent on the initial velocity.
Therefore, if the cannonball is launched at a velocity that matches the orbital velocity, then it will not be able to be drawn in by gravity due to the Earth moving away from the cannonball at the same speed at which the cannonball itself is falling.
This means that the cannonball will continue to fall without reaching the Earth, therefore staying in orbit, much like that of the moon or planets around the sun.
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A 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2 , what is the coefficient of kinetic friction between the box and the inclined plane?
Answer:
Explanation:
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
The coefficient of kinetic friction will be equal to 0.418.
What is friction?Friction is the force that prevents solid surfaces, fluid layers, and material elements from sliding against each other. There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in-touch solid surfaces.
Given that a 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2
The coefficient of kinetic friction will be calculated as,
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
Solve for the value of the coefficient of friction,
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
Therefore, the coefficient of kinetic friction will be equal to 0.418.
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what is the value of acceleration due to gravity g at the pole of earth
Answer:
The value of Acceleration due to gravity at the pole of the earth is 9.870m/s^2 .Explanation:
I hope this helps you !!How do I measure the drag of a paper airplane?
Answer:
hmmmmm ill get back later
Explanation:
CAN SOMEONE PLZ HELP
Answer:
magnetic force.
Explanation:bc it makes sense, and can i please get brainliest answer i never asked. its ok if you say no. have a great day <3.
A student takes the temperature of a bottle of liquid water and a tub filled with ice. The student then places the bottle of liquid water into the tub of ice. After 15 minutes, which of the following best describes what happens to the temperature of each?
There will be exchange of heat in the system, the tub filled with ice will gain heat, while the bottle of liquid will lose heat.
The flow is made possible by means of convection which implies heat flow from region of higher concentration to region of low concentration
by actual movement of material particles
There are basically three modes of heat transfer
RadiationConvectionConductionLearn more:
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Explanation:
if you go to connexus and are doing the science
Thermal Energy Unit Test 10 qestions 8th grd unit 5 test lesson 12 here are the correct answers
1. A. metal is heated from room temperature to 200'C.
2. B. The particles will have more space between them as steam, but they will be moving at the same speed in both states.
3. C. The potential energy decreases due to the tighter arrangement of the particles.
4. B. an increase in heat and an increase in kinetic energy until a phase change occurs.
5. C. thermometer
6. B. 4.
7. D. convection
8. C. Air moves from the areas of higher temperature to areas of lower temperature.
9. C. The temperature of the ice increases, while the temperature of the water decreases
10. A. heat transfer by radiation.
i hope this helps and please let me know if it does! = ) <3
A car slams on its brakes creating an acceleration of -4.7 m/s^2. It comes to rest after traveling a distance of 235 m. What was its velocity before it began to accelerate?
Answer:
Explanation:
v² = u² + 2as
0² = u² + 2(-4.7)(235)
u² = 2209
u = 47 m/s
Help me outttt pls and thanks
