Answer:
a force is represented by a vector the choice of a reference frame is necessary
Explanation:
Correct answer is by. The equiv 48 - (-12) =
Answer:
60
Explanation:
48-(-12)=60
the -(- is like a big plus sign really if you see that then just add the numbers together so do 48 plus 12 it equals 60.
a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf
Answer:
is it a poem or recipe :(
don't know
A small bar of pure gold whose density is 19.3g/cm. Displaces 80 cm
Answer:
The mass of the gold bar is 1,544 g
Explanation:
is everyone lowkey turning goth?
Answer:
Yeah it's the new trend lol it's kinda weird
Answer:
it is marked by mystery and ambiguity. If to this fascination for creating tension is added a feeling of fear
Explanation:
do I help you
An airplane travels 250 km due east, then turns and travels 180 km [E 60°S]. Determine the resulting travel displacement for this plane.
Answer:
388.56km[E13degS]
Explanation:
Δdx = Δd1x + Δd2x
Δdx = 250km [E]+ 180sin(60) [E]
Δdx = 405.88 km [E]
Δdy = Δd1y + Δd2y
Δdy = 0 + 160cos(60) [S]
Δdy = 90km [S]
Δdt = √(Δdx)^2+(Δdy)^2
Δdt = √164738.57+8100
Δdt = 398.85km
tanθ = Δdy/Δdx
tanθ = 90/388.56
θ = tan-1(90/388.56)
θ = 13.04 deg
Which statement describes the law of conservation of energy?
A. There is only one form of energy.
B. When an energy transformation happens, no energy is destroyed
or created.
C. The total energy in a system can only increase.
D. Energy can only change from nuclear energy to chemical energy.
SUBMIT
Answer:
B. When an energy transformation happens, no energy is destroyed
or created.
Explanation:
A 1200-kg car is moving at 17.0 m/s due north. A 1600-kg car is moving at 21.0 m/s due east. The two cars simultaneously approach an icy intersection where, with no brakes or steering, they collide and stick together.
1)
Determine the speed of the combined two-car wreck immediately after the collision. (Express your answer to two significant figures.)
2)
Determine the direction of the combined two-car wreck immediately after the collision. (Express your answer to two significant figures.)
Answer:
f
Explanation:
The speed of the two-car wreck right after the crash is 13.9 m/s, and the direction is 58.4° east of north.
Momentum conservation can address this problem. Before and after the collision, momentum is equal. Calculate the two-car wreck's speed and direction after the collision:
Step 1: Calculate automobile beginning momentum.
Mass (m) times velocity (v) equals momentum (p).
Northbound 1200-kg car:
Initial momentum of the 1200-kg automobile (p₁) = mass (m₁) × velocity (v₁).
p₁ = 1200 kg × 17.0 m/s = 20,400 kg/s (North direction).
1600-kg eastbound car:
The 1600-kg car's initial momentum (p₂) = m₂ × v₂.
p₂ = 1600 kg × 21.0 m/s = 33,600 kg/s (East direction).
Step 2: Calculate the pre-collision momentum.
The vector sum of the two automobiles' momentum before the accident is p[tex]_total[/tex].
p[tex]_total[/tex] = p₁+p₂.
p[tex]_total[/tex] = 20,400 kg m/s (North) + 33,600 (East)
Step 3: Calculate the two cars' combined mass after the collision.
Combined mass (m[tex]_combined[/tex]) = 1200-kg + 1600-kg cars.
m[tex]_combined[/tex] = 1200 + 1600 = 2800 kg
Step 4: Determine the two-car wreck's velocity after the accident.
Total momentum after the collision (p[tex]_combined[/tex]) is the same as before, but it belongs to the combined mass (m[tex]_combined[/tex]) travelling at the velocity (v[tex]_combined[/tex]) in a given direction:
p[tex]_combined[/tex] = m[tex]_combined[/tex] × v[tex]_combined[/tex]
Since the collision is inelastic (cars stick together), momentum remains the same.
= p[tex]_total[/tex]
m[tex]_combined[/tex] × v[tex]_combined[/tex] = 20,400 kg/s (North) + 33,600 kg/s (East).
Step 5: Determine the two-car wreck's speed and direction.
Calculating the resultant vector of the North and East momentum components gives the combined two-car wreck's velocity (speed) (v[tex]_combined[/tex]).
(v[tex]_combined[/tex][tex]_north²[/tex] + v[tex]_combined[/tex][tex]_east²[/tex])
Where:
North velocity is v[tex]_combined[/tex][tex]_north[/tex].
East velocity is v[tex]_combined[/tex][tex]_east[/tex].
v[tex]_combined[/tex][tex]_north[/tex] = 20,400 kg m/s, 2800 kg, 7.29 m/s (North).
v[tex]_combined[/tex][tex]_east[/tex] = 33,600 kg m/s < 2800 kg × 12.0 m/s (East).
13.9 m/s = (7.29² + 12.0²)
Step 6: Determine the two-car wreck's direction.
Trigonometry can determine direction:
tan() = v[tex]_combined[/tex][tex]_east/north[/tex]
arctan(v[tex]_combined[/tex][tex]_east/north[/tex]) =
arctan(12.0 / 7.29 m/s) arctan(1.645) 58.4°
The two-car crash faces east at 58.4°.
Results summary:
The two-car disaster shortly after the accident is 13.9 m/s.
The two-car crash lies 58.4° east of north following the collision.
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A student is planning an experiment to find
out how the height from which he drops a ball
affects how high the ball bounces. What is the control variable?
a) The ball used
b) The height the ball bounces
c) Height of the ball being dropped
Answer:
b) the height the ball bounces
Explanation:
the control variable is the variable that you change yourself. since you change the height that the ball bounces from we know this is the answer
A _ system is one in which one subsystem provides services to another subsystem.
A large system is one in which one subsystem provides services to another subsystem.
What is a large system?This is a type of system that contains many subsystems. The subsystem provides independent services to the larger by communicating, or interacting with each other.
What is a subsystem?A subsystem is smaller units of a large system, designed to perform specific task in the large system.
Thus, a large system is one in which one subsystem provides services to another subsystem.
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In a five paragraph essay, the ________ is the main idea you're trying to prove.
If you were in a spaceship travelling at 0.80 c away from a star, at what speed would the star's light pass you?
Still travelling at 0.80 c you see another spaceship approaching you, and the star. You measure its speed to be 0.84 c.
What is the other spaceship's speed in the star's frame of reference?
(1) The speed with which the star's light will pass you is 3 x 10⁸ m/s.
(2) The speed of the other spaceship is -2.943 x 10⁸ m/s.
The given parameters;
speed of the start light, u' = cspeed of spaceship, v = 0.8c(1)
Since you are travelling away from the star (same direction), relative to a third observer, the speed at which the star's light will pass you is calculated using Lorentz velocity addition.
[tex]u = \frac{u'+ v}{1 + \frac{v}{c^2} u'} \\\\u = \frac{c+(-0.8c)}{1 + \frac{-0.8c}{c^2} (c)}\\\\u = \frac{c-0.8c}{1- \frac{0.8c}{c} } \\\\u = \frac{0.2c}{0.2} = c = 3\times 10^{8} \ m/s[/tex]
Thus, the speed with which the star's light will pass you is 3 x 10⁸ m/s.
(2) The speed of the spaceship approaching = 0.84 c
Since you are travelling in opposite direction to second spaceship, relative to a third observer (frame of reference), the speed of the other spaceship is calculated using Lorentz velocity addition.
let the speed of the approaching spaceship u', = 0.84c
[tex]u = \frac{u' + v}{1 + \frac{v}{c^2}u' } \\\\u = \frac{-0.84c - 0.8c}{1 + \frac{(-0.8c)}{c^2}(-0.84c) } \\\\u = \frac{-1.64c}{1.672} = -0.981c \\\\u = -0.981 \times 3\times 10^{8} = -2.943 \times 10^{8} \ m/s[/tex]
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One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
While your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
Assume both snowballs are thrown with
the same initial speed 15.1 m/s. The first
snowball is thrown at an angle of 69◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? Note the starting and
ending heights are the same. The acceleration
of gravity is 9.8 m/s^2
How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.
Answer:
Range formula: R = v^2 sin (2 theta) / g
If theta = 69 deg and v = 15.1
R = 15.1^2 sin 138 / 9,8 = 15.6 m
sin 138 = .669 = sin 42
So a snowball thrown at 21 deg will travel
R = 15.1 * .669^2 / 9.8 = 15.6 m
The second snowball can be thrown at 21 deg to travel the same distance
Vx = V cos theta = 15.1 * cos 69 = 5.41 first snowball
t1 = 15.6 / 5.41 = 2.88 sec
Vx = V cos theta = 15.1 cos 21 = 14.1 m/s
t2 = 15.6 / 14.1 = 1.11 sec
Difference = t1 - t2 = 1.77 sec time delay for second snowball
What is the tension in the string?
(I’m really having a hard time understanding this topic, and my professor isn’t helping much, so any extra explanation are greatly appreciated). The angles are 54.9205°.
If you look at the sketch I drew for the earlier part of this question, you'll see that, with respect to the positive horizontal (i.e. directly to the right), T₁ makes an angle of 180° - 54.9205° ≈ 125°, while T₂ makes an angle of 54.9205° ≈ 55°.
Split up the force acting on the block into vertical and horizontal components. We have
• net vertical force
∑ F = T₁ sin(125°) + T₂ sin(55°) - mg = 0
where m = 0.56 kg, and
• net horizontal force
∑ F = T₁ cos(125°) + T₂ cos(55°) = 0
Both net forces are 0 because the block is suspended in equilibrium.
Notice that cos(125°) = -cos(55°), so the second equation tells you that T₁ = T₂ and that the tensions in either string are the same. Also, sin(125°) = sin(55°).
Then in the first equation, we have
T₁ sin(125°) + T₂ sin(55°) - mg = 0
2 T₁ sin(55°) = mg
T₁ = mg/(2 sin(55°))
T₁ = (0.56 kg) (9.8 m/s²)/(2 sin(55°))
T₁ ≈ 3.35 N
Which group of the periodic table consists of elements that share similar
properties and have a single electron in their outer shells?
A. 18
B. 2
C. 13
D. 1
Answer:
c 13
Explanation:
I took test
Group 18 of the periodic table, also known as the noble gases or Group 18 elements, consists of elements that share similar properties and have a single electron in their outer shells. The correct option is option (A).
These elements include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). They are characterized by having full outer electron shells, making them stable and unreactive under normal conditions.
This stability is due to the fact that they have achieved a full complement of electrons in their outermost energy level, except for helium, which has only two electrons in total.
Group 18 of the periodic table, also known as the noble gases or Group 18 elements, The correct option is option (A).
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How far will a bus travel if it averages a speed of 65 km/h for 7 hours?
Answer:
the bus travles 65x7=455 km
Explanation:
hope this helps you
Answer:
455 km/h
Explanation:
65km/h x 7 hours = 455 km/h
A racket causes a tennis ball to rebound at a speed of 25 m/s The tennis ball is 0.4 kg and is initially moving at a speed of 12 m/s. A high-speed movie film determines that the racket and ball are in contact for 0.05 s. What is the average net force exerted on the ball by the racket
Hi there!
We know that:
I = Δp = mΔv = m(vf - vi)
Let the positive direction be towards the wall. Plug in the given values:
Δp = 0.4(-25 - 12) = -14.8 Ns
Δp = Ft, so:
Δp/t = F
-14.8/0.05 = -296 N (depending on which direction you assign positive/negative, the answer would be positive/negative)
A car initially traveling at a speed of 15.0 m/s accelerates uniformly to a speed of 20.0 m/s over a distance of 40.0 meters. What is the magnitude of the car's acceleration?
Answers:
1.1 m/s^2
2.0 m/s^2
2.2 m/s^2
9 m/s^2
Answer:
[tex]\boxed {\boxed {\sf 2.2 \ m/s^2}}[/tex]
Explanation:
We are asked to solve for the magnitude of the car's acceleration.
We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,
[tex]v_f[/tex]= 20.0 m/s[tex]v_i[/tex]= 15.0 m/s d= 40.0 mSubstitute the values into the formula.
[tex](20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)[/tex]
Solve the exponents.
(20.0 m/s)² = 20.0 m/s * 20.0 m/s = 400.0 m²/s² (15.0 m/s)² = 15.0 m/s * 15.0 m/s = 225.0 m²/s²[tex]400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)[/tex]
Subtract 225.0 m²/s² from both sides of the equation.
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)[/tex]
[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)[/tex]
[tex]175 \ m^2/s^2 = 2a(40.0 \ m)[/tex]
Multiply on the right side of the equation.
[tex]175 \ m^2/s^2 =80.0 \ m *a[/tex]
Divide both sides by 80.0 meters to isolate the variable a.
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}[/tex]
[tex]\frac {175 \ m^2/s^2}{80.0 \ m}=a[/tex]
[tex]2.1875 \ m/s^2 =a[/tex]
Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.
[tex]2.2 \ m/s^2=a[/tex]
The magnitude of the car's acceleration is 2.2 meters per second squared.
Una varilla de 5m de longitud y 1.5 cm^2 de sección transversal se alarga 0.10 cm al someterla a una tensión de 700 N. Determinar el Módulo de Young de la varilla
What is the rotational speed of the hour hand on a clock?
Answer:
π/21600 rad /s.
a boat travel due East with at 40metre per second across a river flowing due South 30metre per Second. what is the resultant speed of the boat
Answer:
Explanation:
Use vector addition
v = √(30² + 40²) = 50 m/s as viewed from the shore
Trust me when I say I would never venture onto a river moving at 30 m/s no matter how powerful the boat was.
30 m/s(3600 s/hr / 1000 m/km) = 108 km/hr
The turbulence in water flowing over ground at 108 kilometers per hour would capsize anything ever built...including aircraft carriers.
These numbers are more suitable for a plane flying in air. Even that could be a very rough ride.
Study the venn diagram given below carefully. Which of the following can be placed at B? Options: Clay _ Paper _ Water_ Nitrogen.
Answer:
water and nitrogen can be place at B
The B loop is the element that can be liquid, cas or solid. Water can have three states : solid (ice), liquid, or gas (as a vapor). The same can be said with nitrogen, it's a gas at room temperature, and it also can be a liquid or solid.
Two cars drive on a straight highway. At time t = 0 , car 1 passes ad marker 0 traveling due east with a speed of 20.0 m/s. At the same time, car 2 is 1.0 km east of road marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up, with an acceleration of 2.5 m/s², and car 2 is slowing down, with an acceleration of -3.2 m/s².
(a) Write position-time equations for both cars. Let east be the positive direction.
(b) At what time do the two cars meet?
Answer: (a) Write position-time equations for both cars. Let east be the positive direction.
Explanation:
Hope this helps!
An object with an initial velocity of 2 m/s moves in a straight line under a constant acceleration.
Three seconds later, its velocity is 16 m/s.
(a) How far did the object travel during this time?
(b) What was the acceleration of the object?
Answer:
(a) distance =27m
(b)acceleration =14/3ms-2
Explanation:
s=(u+v)t/2
= (2+16)3/2
=27m
v=u+at
16=2+a*3
a= 14/3ms-2
What happens to the direction of the line joining when the object slows down ?Explain the observations
Answer:
it will be curved as in deceleration
Explanation:
A 5000 kg train is traveling at a velocity of 100 m/s and hits another train. The two trains stick together and the new velocity is 50m/s. What is the mass of the second train?
Answer:
Explanation:
Conservation of momentum is used to solve
Unfortunately we have a missing piece of information such as the initial velocity of the unknown mass train.
If we ASSUME that the second train is at rest
5000(100) + m(0) = 5000(50) + m(50)
which means m = 5000 kg
However, I'll show you the importance of knowing that initial velocity by finding it assuming the other answers are valid
if m = 15000 kg
5000(100) + 15000(v₀) = (5000 + 15000)(50)
v₀ = 33 ⅓ m/s
if m = 10000 kg
5000(100) + 10000(v₀) = (5000 + 10000)(50)
v₀ = 25 m/s
if m = 8000 kg
5000(100) + 8000(v₀) = (5000 + 8000)(50)
v₀ = 18.75 m/s
So you can see why I had to assume an initial velocity. Any of the masses could work if the initial velocity is chosen correctly.
How can people help stop erosion on beach
Answer:
since erosion is unavoidable the problem becomes discovering ways to prevent it. present beach erosion prevention methods include sand bags,vegetation,seawalls,sand dunes,and sand fences.How to Convert 20000N/m2 to N/cm2 ? Explain it Step by Step.
Answer:
2N/cm²
Explanation:
1m=100cm so, 1m²= 100×100 cm² = 10000cm².
therefore , 20000N/m²= 20000/10000 N/cm².
= 2N/cm².
hope this helps you.
an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed
Answer:
Explanation:
ASSUMING the belt is horizontal
kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N
Horizontal acceleration is
a = F/m = 47.04 / 8 = 5.88 m/s²
t = v/a = 5.0 / 5.88 = 0.85034...
t = 0.85 s
21)
uses a hypothesis to describe the relationship between dependent and independent variables and
results in advances in scientific knowledge.
A)
Engineering method
1
B)
Architectural design
Eliminate
C)
Technological design
D
Scientific investigation
im not sure if im right or not but maybe its "D scientific investigation" but ldk so
Flag An alpha particle (consisting of two protons and two neutrons) is moving in a circle at a constant speed, perpendicular to a uniform magnetic field applied by some current-carrying coild. The alpha particle makes an clockwise revolution every 88 ns. If the speed is small compared to the speed of light, what is the magnitude B of the magnetic field made by the coils
Answer:
m = 4 At Mass units = 4 * 1.66E-27 kg = 6.64E-27 kg mass of particle
q = 2 * 1.60E-19 C = 3.2E-19 C total charge of alpha particle
T = 8.8E-8 s period of 1 revolution
F = q V B magnetic force
F = m V^2/ R centripetal acceleration
q V B = m V^2 / R
B = m V / (q * R)
T = 2 pi R / V time for 1 revolution
R = V T / (2 pi)
B = (m V / q) * 2 pi / (V T) = 2 pi m / (q T)
B = 2 pi * 6.64E-27 / (3.2E-19 * 8.8E-8) = 1.48
B = 1.48 W/m^2