Answers
Answer:
(a) 12 m/s
(b) At t = 0, x'(t) = 0
At t = 5.0 s, x'(t) = 15 m/s
At t = 10.0 s, x'(t) = 12 m/s
(c) i. The initial velocity = 2.0 cm/s
The initial position = 50 cm
The initial acceleration = 0.125 cm/s²
ii. 16 seconds
2. (a) Approximately 0.85 m/s
(b) 2.094 J
(c) i. Approximately 49.74 m/s
ii. Approximately 86.38°
3. (i) 4·i + 5·j
(ii) -2·i - j - 2·k
(iii) 5·i + 4·j - 3·z
(iv) 8
(v) (a) The magnitude is 2.8 cm, the direction is East
(b) The magnitude is ((14·√3)/5) cm, the direction is North
(c) The magnitude is ((14·√3)/5), the direction is South
Explanation:
x(t) is given as follows;
x(t) = b·t² - c·t³
Where;
b = 2.4 m/s(²) (we note that the unit of b for the term to be distance in m/s²)
c = 0.120 m/s³
(a) At t = 0, we have;
x(0) = b × 0² - c × 0³ = 0
At t = 10.0 s, we have;
x(10.0) = 2.4 m/s² × (10.0 s)² - 0.120 m/s³ × (10.0 s)³ = 120 m
The average velocity = (Total distance)/(Total time)
∴ The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex], is given as follows;
[tex]v_{ave}[/tex] = (120 m - 0 m)/(10.0 s - 0 s) = 12 m/s
The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex] = 12 m/s
(b) The instantaneous velocity, [tex]v_{inst}[/tex], is given as follows;
[tex]v_{inst} = \lim \limits_{t \to 0} \left( v_{ave}\right) = \lim \limits_{t \to 0} \left( \dfrac{\Delta x}{\Delta t} \right) = \dfrac{dx}{dt} = x'(t)[/tex]
[tex]x'(t) = \dfrac{d\left (b \cdot t^2 - c \cdot t^3\right)}{dt}[/tex]
x'(t) = 2·b·t - 3·c·t²
At t = 0, x'(t) = 2 × b × 0 - 3 × c × 0² = 0
At t = 5.0 s, x'(t) = 2 × 2.4 m/s² × 5.0 s - 3 × 0.120 m/s³ × (5.0 s)² = 15 m/s
At t = 10.0 s, x'(t) = 2 × 2.4 m/s² × 10.0 s - 3 × 0.120 m/s³ × (10.0 s)² = 12 m/s
(c) x(t) = 50 cm + (2.0 cm/s)·t - (0.0625 cm/s²)·t²
i. The initial velocity is the instantaneous velocity, x'(t), at time, t = 0
x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t
At t = 0, x'(0) = 2.0 cm/s - 2 × 0.0625 cm/s² × 0 = 2.0 cm/s
The initial velocity, x'(0) = 2.0 cm/s
The initial position = The position at time t = 0 = x(0)
x(0) = 50 cm + (2.0 cm/s) × 0 - (0.0625 cm/s²) × 0² = 50 cm
The initial position, x(0) = 50 cm
The initial acceleration, x''(0) = 2 × 0.0625 cm/s² = 0.125 cm/s²
ii. x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t
When the velocity of the turtle, x'(t) = 0 we have;
0 = 2.0 cm/s - 2 × 0.0625 cm/s² × t
∴ t = (2.0 cm/s)/(2 × 0.0625 cm/s²) = 16 seconds
The velocity of the turtle is zero after 16 seconds
2. The mass of the large fish, m₁ = 15.0-kg
The speed of the large fish, v₁ = 1.1 m/s
The mass of the smaller fish, m₂ = 4.50 kg
The speed of the small (stationary) fish, v₂ = 0
The initial momentum = 15.0 kg × 1.1 m/s + 4.50 kg × 0 = 16.5 kg·m/s
The initial momentum = 16.5 kg·m/s
The final momentum = (15.0 kg + 4.50 kg) × v₃ = 19.50 kg × v₃
The final momentum = 19.50 kg × v₃
Where;
The total initial momentum = The total final momentum
We get;
16.5 kg·m/s = 19.50 kg × v₃
∴ v₃ = (16.5 kg·m/s)/(19.50 kg)
v₃ = (16.5/19.50) m/s = (11/13) m/s ≈ 0.85 m/s
∴ The speed of the large fish just after it eats the small, v₃ ≈ 0.85 m/s
(b) The initial kinetic energy, K.E.₁ = (1/2) × 15 kg × (1.1 m/s)² = 9.075 J
The final kinetic energy, K.E.₂ = (1/2) × 19.5 kg × (11/13 m/s)² = 363/52 J
The mechanical energy dissipated, ΔE = K.E.₁ - K.E.₂
ΔE = (9.075 - 363/42) J = 1089/520 J ≈ 2.094 J
The mechanical energy dissipated, ΔE = 2.094 J
(c) i. We have the total momentum = 110 × 8.8· j + 85 × 7.2· i = 9.680·i + 612·i
The velocity after collision, v = (9.680·i + 612·i)/(110 + 85) = 49.64·j + 3.14·i
The magnitude of the velocity, v = √(49.64² + 3.14²) ≈ 49.74 m/s
ii. The direction, θ = arctan(49.64/3.14) ≈ 86.38°
3. (i) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) + 3·i + 3·j + k = 4·i + 5·j
(ii) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) - (3·i + 3·j + k) = -2·i - j - 2·k
(iii) [tex]\underset{A}{\rightarrow} \times \underset{B}{\rightarrow}[/tex] = (2 + 3)·i - (1 + 3)·j + (3 - 6)·z = 5·i + 4·j - 3·z
(iv) [tex]\underset{A}{\rightarrow} \cdot \underset{B}{\rightarrow}[/tex] = 1×3 + 2 × 3 + (-1)×1 = 8
(v) (a) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j + (2.8·cos(60°)·i - 2.8·sin(60°)·j
[tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 5.6·cos 60°·i = 2.8·i
The magnitude = 2.8, the direction is east
(b) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j - (2.8·cos(60°)·i - 2.8·sin(60°)·j
[tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 5.6·sin(60°)·j= ((14·√3)/5)·j
The magnitude = ((14·√3)/5), the direction is North
(c) [tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = (2.8·cos(60°)·i - 2.8·sin(60°)·j - (2.8·cos(60°)·i + 2.8·sin(60°)·j)
[tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = -5.6·sin(60°)·j= (-(14·√3)/5)·j
The magnitude = ((14·√3)/5), the direction is South
Related Questions
if the action force is 100N what will be the reaction force
Answers
Answer:
HONORS PHYSICS
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Silly Beagle
Newtons's 3rd Law of Motion
Newton’s 3rd Law of Motion, commonly referred to as the Law of Action and Reaction, describes the phenomena by which all forces come in pairs. If Object 1 exerts a force on Object 2, then Object 2 must exert a force back on Object 1. Moreover, the force of Object 1 on Object 2 is equal in magnitude, or size, but opposite in direction to the force of Object 2 on Object 1. Written mathematically:
Newton's 3rd Law Equation
This has many implications, some of which aren’t immediately obvious. For example, if you punch the wall with your fist with a force of 100N, the wall imparts a force back on your fist of 100N (which is why it hurts!). Or try this. Push on the corner of your desk with your palm for a few seconds. Now look at your palm... see the indentation? That’s because the corner of the desk pushed back on your palm.
running tiger
Although this law surrounds your actions everyday, often times you may not even realize its effects. To run forward, a cat pushes with its legs backward on the ground, and the ground pushes the cat forward. How do you swim? If you want to swim forwards, which way do you push on the water? Backwards, that’s right. As you push backwards on the water, the reactionary force, the water pushing you, propels you forward. How do you jump up in the air? You push down on the ground, and it’s the reactionary force of the ground pushing on you that accelerates you skyward!
As you can see, then, forces always come in pairs. These pairs are known as action-reaction pairs. What are the action-reaction force pairs for a girl kicking a soccer ball? The girl’s foot applies a force on the ball, and the ball applies an equal and opposite force on the girl’s foot.
How does a rocket ship maneuver in space? The rocket propels hot expanding gas particles outward, so the gas particles in return push the rocket forward. Newton’s 3rd Law even applies to gravity. The Earth exerts a gravitational force on you (downward). You, therefore, must apply a gravitational force upward on the Earth!
Which statement correctly describes the Earth's magnetic field?
A. The geomagnetic south is consistent with a magnetic south.
B. The geomagnetic north is similar to the south pole of a magnet.
C. The Earth's magnetic poles are the same magnetic charge at the poles.
D. The Earth does not have a consistent pattern of magnetism.
Answers
Answer: B. The geomagnetic north is similar to the south pole of a magnet
Explanation:
HELPPP I NEED HELP ASAP NOW
Answers
Answer:
Your answers would be
4. A. sperm and testosterone.
7. C. prostate, penis, Testes
uterus, vagina, fallopian tubes
10. B. Protein
11. A. carbohydrate
12. B. amino acids (I'm not positive on this i haven't taken bio in years
27. D. Respiratory system
Explanation:
yeah
A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block
of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2.
After the collision the bullet becomes embedded in the block.
A) What is the net momentum of the bullet-block system before the collision?
B) Find the total energy of the bullet-block system before the collision?
C) What is the speed of the bullet-block system after the collision?
D) *Find the total energy of the bullet-block system after the collision?
E) *How much work must be done to stop the bullet-block system?
F) *Find the maximum traveled distance of the bullet-block after the collision?
Answers
Answer:
a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is 89.289 joules.
e) 89.289 joules must be done to stop the bullet-block system.
f) The bullet-block system will travel 13.007 meters before stopping.
Explanation:
a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by the initial momentum of the bullet:
[tex]p = m\cdot v_{o}[/tex] (1)
Where:
[tex]p[/tex] - Net momentum, in kilogram-meters per second.
[tex]m[/tex] - Mass of the bullet, in kilograms.
[tex]v_{o}[/tex] - Initial speed of the bullet, in meters per second.
If we know that [tex]m = 0.5\,kg[/tex] and [tex]v_{o} = 50\,\frac{m}{s}[/tex], then the net momentum of the bullet-block system before the collision is:
[tex]p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]p = 25\,\frac{kg\cdot m}{s}[/tex]
The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The total energy of the bullet before the collision is its initial translational kinetic energy ([tex]K[/tex]), in joules:
[tex]K = \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex] (2)
[tex]K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 625\,J[/tex]
The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:
[tex]v_{f} = \frac{m\cdot v_{o}}{m+M}[/tex] (3)
Where:
[tex]v_{f}[/tex] - Final speed, in meters per second.
[tex]M[/tex] - Mass of the block, in kilograms.
If we know that [tex]m = 0.5\,kg[/tex], [tex]v_{o} = 50\,\frac{m}{s}[/tex] and [tex]M = 3\,kg[/tex], then the final speed of the bullet-block system is:
[tex]v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]v_{f} = 7.143\,\frac{m}{s}[/tex]
The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system ([tex]K[/tex]), in joules, is:
[tex]K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2}[/tex] (4)
[tex]K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 89.289\,J[/tex]
The total energy of the bullet-block system after the collision is 89.289 joules.
e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.
f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:
[tex]W = \mu_{k}\cdot (m+M)\cdot g\cdot s[/tex] (5)
Where:
[tex]W[/tex] - Work done by friction, in joules.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
If we know that [tex]m = 0.5\,kg[/tex], [tex]M = 3\,kg[/tex], [tex]\mu_{k} = 0.2[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]W = 89.289\,J[/tex], then the travelled distance of the bullet-block system is:
[tex]s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}[/tex]
[tex]s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 13.007\,m[/tex]
The bullet-block system will travel 13.007 meters before stopping.
QUESTION 30 A tennis ball moves back and forth 10 times in 5 sec. The frequency of its motion is
Answers
Answer:
so in 1 sec 2 times
so frequency = 2
Explanation:
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answers
Answer:
[tex]F=0N[/tex]
Explanation:
From the question we are told that:
Charge [tex]Q=4.88 x 10-6 C[/tex]
Velocity [tex]v= 265m/s[/tex]
Angle [tex]\theta =0 \textdegree[/tex]
Magnetic field [tex]B=0.0579T[/tex]
Generally the equation for Force is mathematically given by
[tex]F=Q(\=v*\=B)[/tex]
[tex]F=qvBsin\theta[/tex]
Therefore
[tex]F=qvBsin0 \textdegree[/tex]
[tex]F=0N[/tex]
Answer:
0 newtons
Explanation:
When a moving charge is parallel to the magnetic field, it feels no Magnetic force at all.
Even if the Magnetic Field is 100,000,000 Tesla!
Convert:
1) 2kg into gram
2) 5200m into km
3) 20cm into m
Answers
2) 5200 meters converted into kilometers is 5.2
3) 20 centimeters converted into meters is 0.2
Why is a person not a good blackbody radiator?
O A. A person emits only visible light.
OB. A person emits only infrared radiation.
O C. A person absorbs most of the light that hits him or her.
O D. A person reflects little of the light that hits him or her.
Answers
Answer:
O C. A person absorbs most of the light that hits him or her.
Explanation:
Answer:
Option D hope it's helpful mark me as brainlistEm um fio condutor uma carga de 6.000 C atravessa uma secção transversal em 5 minutos. Determinando-se a corrente no fio, encontraremos o valor de?
Answers
Answer:
I = 20 A
Explanation:
The question says that, "A load of 6,000 C is conducted through a cross section in 5 minutes. Determining-if a current is not correct, we will find the value of?"
We have,
Charge, q = 6,000 C
Time, t = 5 minutes = 300 s
We need to find the current. We know that, the charge flowing per unit time is equal to current. So,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{6000}{300}\\\\I=20\ A[/tex]
So, the current flowing through the circuit is 20 A.
A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What is the ball speed just before hitting the ground?
Answers
Answer:
Explanation:
Here's what we know because it was given to us:
a = -9.8 m/s/s and
time = 3.32 seconds
Here's what we know because we rock physics:
v₀ = 0 (because the object was held still before it was dropped).
Here's the equation that ties all that info together in a single one-dimensional equation:
v = v₀ + at
Filling in and solving for v:
v = 0 + (-9.8)(3.32) and
v = -33m/s
The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 52.1 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block
Answers
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
(Please help if you can, I need this last answer done by tonight.)
Use the universal law of gravitation to solve the following problem.
Hint: mass of the Earth is = 5.97 x 1024 kg
A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?
a. Write out the formula for this problem.
b. Plug in the values from this problem into the formula.
c. Solve the problem, writing out each step.
d. Correct answer
Answers
Answer:
a.
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
d. 1.144 × 10¹³ N
Explanation:
The universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
The given mass of the scientific satellite, m = 1,300 kg
The height of the orbit of the satellite, r = 200 km above the Earth's surface
The length of the radius of the Earth, R = 6378 km
The mass of the Earth = 5.97 × 10²⁴ kg
a. The formula for the universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
Where;
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the satellite = 1,300 kg
r = The distance between the Earth and the satellite = R + r = 6,378 km + 200 km = 6,578 km
G = The Gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
b. Plugging in the values from the problem into the formula gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c. Solving gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
The force acting between the Earth and the satellite, F ≈ 1.144 × 10¹³ N
d. 1.144 × 10¹³ N
Two cars move at different velocities. Car A moves at a speed of 90 km/hr while Car B moves
at a speed of 75 km/hr . If both cars collided with a wall made up of the same material, which of
the two cars will create greater impact of collision. Both cars have the same masses.
(Serious Answers Please)
Answers
This type of collision is known as an Inelastic collision.
The speed rate at which car a used is more higher than that of car b because of its friction.
I hope this helps.
A car takes a full round of journey in a roundabout with constant speed, as the driver got confused with the route. Can we consider it as a uniform motion? Why?
Answers
Answer: The given statement is True
Explanation:
A uniform motion is defined as the motion where an object is moving at a constant speed.
A non-uniform motion is defined as the motion where an object keeps changing its position and does not move at a constant speed.
We are given:
A car takes a full round of journey in a roundabout with constant speed
As the speed remains constant in a circular path, it is considered a uniform motion.
Hence, the given statement is True
distance= 10km due West in 1hour calculate the velocity
Answers
Answer:
Velocity = distance / time
V = 10/1
V = 10km/h
Answer:10km/h or 2.77m/s.
Explanation:
Distance =10km
Time =1h
Velocity =10/1 =10km/h
Or,
Distance =10km =10000m
Time =1h =60min = 3600s
Velocity =10000/3600 =2.77m/s
Explain the energy transformations that occur when accelerating in a gasoline
vehicle.
Answers
A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What is its temperature at this new volume and pressure?
Answers
Answer:
27°C
Explanation:
We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27 °C
Initial temperature (T₁) = 27 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall determine the final temperature of the gas. This can be obtained as follow:
Initial volume (V₁) = 2 m³
Initial temperature (T₁) = 300 K
Initial pressure (P₁) = 1 atm
Final pressure (P₂) = 2 atm
Final volume (V₂) = 1 m³
Final temperature (T₂) =?
P₁V₁/T₁ = P₂V₂/T₂
1 × 2 / 300 = 2 × 1 / T₂
2/300 = 2/T₂
1/150 = 2/T₂
Cross multiply
T₂ = 150 × 2
T₂ = 300 K
Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 300 K
T(°C) = 300 – 273
T(°C) = 27°C
Thus, the final temperature is 27°C
In a bicycle dynamo,does 1. The permanent magnet surrounds a conducting coil 2. The conducting coil rotates when the rear wheel of the bicycle rotates. 3.The electricity is generated in the permanent magnet 4.When the rear wheel of the bicycle turns fast, the brightness of the light increases
Answers
Explanation:
3: the electricity is generated in the permanent magnet
I know this because yea I got this and I got it correct so trust me
Una masa de Hidrogeno ocupa 0.2 L a 100°C. Determine su volumen a 0°C, si la presión se mantiene constante. Como la presión y la cantidad de materia se mantienen constantes, podemos aplicar la ley de Charles
Answers
Answer:
0,146 L
Explanation:
Según la ley de Charles; el volumen de una determinada masa de gas es directamente proporcional a su temperatura a presión constante.
A partir de los datos proporcionados;
V1 = 0,2 L
T1 = 100 ° C + 273 = 373 K
V2 =
T2 = 0 ° C + 273 = 273 K
V1 / T1 = V2 / T2
V1T2 = V2T1
V2 = V1T2 / T1
V2 = 0,2 × 273/373
V2 = 0,146 L
The earth's orbital is oval in shape. Explain how the magnitude of the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure.
Answers
The gravitational force does change believe it or not, but the explaination for this is because the earths orbit is an oval (or a not circle) the closer it nears its self to the sun.
Which of the following could be used to create an open circuit?
Answers
What are the answers choices
A horse pulls a wagon with a force of 200 N for a distance of 80 m. How much work
does the horse do?
Answers
Answer:
w=f×s
w = 16000 J, hope this helps
help....................
Answers
Calculate the man’s mass. (Use PE = m × g × h, where g = 9.8 N/kg.)
A man climbs a wall that has a height of 8.4 meters and gains a potential energy of 4,620 joules. His mass is about
kilograms
Answers
Answer:
mass=56.12kg
Explanation:
PE=mgh
4620=m×9.8×8.4
make m subject of the formula...
m =4620/(9.8×8.4)
m=4620/82.32
m=56.12kg
A converging lens can be defined as __________
a lens that causes parallel light rays to bounce off the surface
a lens that allows parallel light rays to pass without changing direction
a lens that causes parallel light rays to separate from each other
a lens that causes parallel light rays to focus at a specific location
Answers
Answer:
a lens that causes parallel light rays to separate from each other
Answer:
a lens that causes parallel light rays to focus at a specific location
I took the test and got it right! :)
sulfur and oxygen can react to form both sulfur dioxide and sulfur trioxide in sulfur dioxide there are 32.06 grams of sulfur and 32 grams of oxygen in sulfur dioxide there are 32.06 grams of sulfur are combined with 48 grams of oxygen
a. what is the ratio of the weights of oxygen that combine with 32.06 g of sulfur ?
b. How do these data illustrate the law of multiple proportions?
Answers
Answer:
a. 2:3
b. The data illustrates the law of multiple proportions by showing that the the masses of oxygen that reacts with a fixed mass of sulfur are in a ratio of small whole numbers
Explanation:
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur dioxide = 32 grams
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur trioxide = 48 grams
a. The ration of the weights of oxygen that combine with 32.06 g of sulfur = 32:48 = 2:3
b. The law of multiple proportions states that when two elements are able to interact chemically to form more than one compound, then the (different) weights of one of the element that combines with a fixed weight of the other element are in small whole number ratios
The data demonstrates the law of multiple proportions by showing that the ratios of the weights of oxygen that combine with a fixed weight of sulfur to form sulfur dioxide and sulfur trioxide is in the ratio of 2 to 3 which are small whole number ratios
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle invlove and use of which functions
Answers
Answer:
Explanation:
The formula to analyze motion in the vertical (or y) dimension is
[tex]A_y=Asin\theta[/tex] which says that the motion in the y-dimension is equal to the magnitude of the A vector times the sin of the angle.
The formula to analyze motion in the horizontal (or x) dimension is
[tex]A_x=Acos\theta[/tex] which says that the motion in the x-dimension is equal t the mignitude of the A vector times the cosine of the angle.
Many times this is used to find the upwards velocity and the horizontal velocity when an overall velocity is given with an angle of inclination.
When magma flows on the surface on the surface, it is already called lava
TRUE OR FALSE
Answers
Answer:
True
Explanation:
I guess you made a mistake on question.
but I understood what you wanted to say.
Hope this helps... :)
Neha and Reha are playing see-saw.Neha is sitting 60cm away from the fulcrum and Reha is sitting 40cm away from the fulcrum.Calculate the effort that Reha should apply to lift Neha.The weight of Neha is 360N.
Answers
Answer:
Effort = 540 Newton
Explanation:
Given the following data;
Load arm = 60 cm
Effort arm = 40 cm
Load = 360 N
Conversion:
100 cm = 1 meters
40 cm = 40/100 = 0.4 meters
60 cm = 60/100 = 0.6 meters
To calculate the effort that Reha should apply to lift Neha, we would use the expression;
Effort * effort arm = load * load arm
Substituting into the expression, we have;
Effort * 0.4 = 360 * 0.6
Effort * 0.4 = 216
Effort = 216/0.4
Effort = 540 Newton
The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is received with a time delay of 0.50s after the original sound is emitted.
Calculate the distance between submarine and the other object
Answers
Answer:
d = 375 m
Explanation:
The speed of sound is constant in any medium, therefore we can use the uniform motion relationships
v = x / t
x = v t
In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance
x = 2d
2d = v t
d = v t/2
let's calculate
d = 1500 0.5 / 2
d = 375 m