The minimum power is 490 W
Explanation:
The work that needs to be done by the robot in order to climb the shelves is equal to the gain in gravitational potential energy of the robot + object, therefore:
[tex]W=mg \Delta h[/tex]
where
m = 15 kg + 5 kg = 20 kg is the total mass of the system
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\Delta h = 20 m[/tex] is the change in height
Substituting,
[tex]W=(20)(9.8)(20)=3,920 J[/tex]
Now we can calcualte the power (wattage) that the robot must have, using the equation
[tex]P=\frac{W}{t}[/tex]
where
W = 3,920 J is the work done
t = 8 s is the time interval
Substituting,
[tex]P=\frac{3920}{8}=490 W[/tex]
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Can someone explain the gas laws?
* ex...what happens to pressure if volume and temp. increase*
Explanation:
Use the Ideal Gas Law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is the universal gas constant,
and T is absolute temperature.
First, identify which variables in the problem are constant (note that R is always a constant).
Next, solve for the constants and use that equation to write a proportion.
Finally, plug in the given values and solve for the unknown variable.
For example, if n and R are the constants:
nR = PV/T
Therefore:
P₁V₁/T₁ = P₂V₂/T₂
A 1500kg car, moving at a speed of 20m/s comes to a halt. How much work was done by the brakes?
The work done by the brakes is [tex]-3.0\cdot 10^5 J[/tex]
Explanation:
According to the work-energy theorem, the work done by the brakes on the car is equal to the change in kinetic energy of the car. Therefore:
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done
m is the mass of the car
v is the final speed
u is the initial speed
For the car in this problem, we have:
m = 1500 kg
u = 20 m/s
v = 0 (the car comes to a halt)
Substituting, we find the work done:
[tex]W=0-\frac{1}{2}(1500)(20)^2=-3.0\cdot 10^5 J[/tex]
And the work is negative because the brakes apply a force in the opposite direction to the motion of the car.
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What is Snell law of refraction
Answer:
Snell's law tells us the degree of refraction and relation between the angle of incidence, the angle of refraction and refractive indices of given pair of media. We know that light experiences the refraction or bending when it travels from one medium to another medium. Snell's law predicts the degree of the bend.
Explanation:
Final answer:
Snell's law expresses the relationship between the incidence and refraction angles and the indices of refraction of two media, represented by the equation n₁ sin θ₁ = n₂ sin θ₂.
Explanation:
Snell's law describes the relationship between angles and indices of refraction when light passes from one transparent medium to another.
It is an essential concept in the study of optics and can be mathematically expressed as n₁ sin θ₁ = n₂ sin θ₂, where n₁ and n₂ are the refractive indices of the first and second medium, respectively, and θ₁ and θ₂ are the angles of incidence and refraction.
This law is named after the Dutch mathematician Willebrord Snell who described it in 1621, but it was also discovered by the Arabian physicist Ibn Sahl in the 10th century.
Through thorough experiments, it became evident that light rays change direction due to the differing speeds of light in various media, a concept Snell himself was not aware of but was able to confirm through the refractive indices.
if you are traveling at 75 km/h how long will it take to travel 32 km?
The time taken is approximately 26 minutes
Explanation:
The motion of the body in this problem is a uniform motion (= at constant velocity), therefore we can use the following equation:
[tex]v=\frac{d}{t}[/tex]
where
v is the speed
d is the distance covered
t is the time taken
In this problem, we have
v = 75 km/h is the speed
d = 32 km is the distance to be covered
Solving for t, we find the time needed:
[tex]t=\frac{d}{v}=\frac{32}{75}=0.43 h[/tex]
Converting into minutes,
[tex]t=0.43 h \cdot 60 = 25.8 min \sim 26 min[/tex]
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A truck accelerates to a velocity of 38 m/s over 755 m of road
1) The acceleration is [tex]0.96 m/s^2[/tex]
2) The time taken is 39.6 s
Explanation:
1)
Since the motion of the truck is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the truck in this problem,
u = 0 (it starts from rest)
v = 38 m/s
s = 755 m
Solving for a, we find the acceleration:
[tex]a=\frac{v^2-u^2}{2s}=\frac{38^2-0}{2(755)}=0.96 m/s^2[/tex]
2)
For this part we can use the following suvat equation
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken for the velocity to change from u to v
In this problem,
u = 0
v = 38 m/s
[tex]a=0.96 m/s^2[/tex]
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{38-0}{0.96}=39.6 s[/tex]
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The truck's acceleration, starting from rest to a velocity of 38 m/s over 755 m, is approximately 0.956 m/s². It takes about 39.75 seconds to cover this distance.
a. To determine the acceleration of the truck, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (38 m/s),
u = initial velocity (0 m/s since the truck was initially at rest),
a = acceleration,
s = displacement (755 m).
Rearranging the equation to solve for acceleration (a), we get:
a = (v^2 - u^2) / (2s)
Substitute the given values:
a = (38^2 - 0^2) / (2 * 755)
a = 1444 / 1510
a ≈ 0.956 m/s^2
b. To find the time taken (t), we can use the kinematic equation:
v = u + at
Since the truck started from rest (u = 0), this simplifies to:
t = v / a
Substitute the values:
t = 38 / 0.956
t ≈ 39.75 s
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The question probable may be:
A Truck accelerates to a velocity of 38 m/s over 755 m of road.
a. If the truck was initially at rest, what was its acceleration?
b. How long willthe truck take to travel this distance?
A roller coaster engineer is trying to design a new coaster to build at the theme park. The cars, when loaded with people, is expected to have a mass of 8000. He wants his roller coaster cars to be going 10 m/s when it reaches the bottom. Assuming that friction is negligible, how high up should the roller coaster start?
_________Number ___M___Units
Then Answer
Well, it turns out that friction is not negligible, the Engineer figured out that the car is going to lose 32000 Joules energy as it goes down the track. this means that he has to start up even higher. what is the actual starting height after taking into account the energy loss from friction?
__________Number _____M____Units
Thank you SOO much :D
1) The initial height of the roller coaster must be 5.1 m
2) The height of the roller coeaster must be 5.5 m
Explanation:
1)
We can solve this problem by using the law of conservation of energy.
In fact, in absence of frictional force, the total mechanical energy must be conserved, so we can write:
[tex]KE+PE = KE'+PE'[/tex]
where
KE = 0 is the initial kinetic energy, at the top (the car starts from rest, so it has zero speed and so zero kinetic energy)
PE = mgh is the initial potential energy, where
m = 8000 kg is the mass of the car
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h is the initial height
[tex]KE'=\frac{1}{2}mv'^2[/tex] is the kinetic energy of the car at the bottom, where
v' = 10 m/s is the speed of the car at the bottom
PE' = 0 is the potential energy at the bottom (where h = 0)
Re-writing the equation with all the values and solving for h, we find the initial height of the track:
[tex]h=\frac{v^2}{2g}=\frac{10^2}{2(9.8)}=5.1 m[/tex]
2)
In this case, we also have to take into account the energy lost due to friction. So the equation of conservation of energy becomes
[tex]KE+PE = KE'+PE'+W[/tex]
where
W = 32,000 J is the work done by friction
Re-arranging the equation and solving again for h, we find:
[tex]mgh = \frac{1}{2}mv'^2 + W\\h=\frac{v^2}{2g}+\frac{W}{mg}=5.1+\frac{32,000}{(8000)(9.8)}=5.5 m[/tex]
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Most of the Earth's volume is contained in the
A. mantle
B.
crust
C.
inner core
D.
outer core
Most of the Earth's volume is contained in the "mantle". The correct answer would be option (A).
What is a mantle?The mantle is a layer of semi-solid rock that lies between the Earth's thin crust and its liquid outer core. The mantle is approximately 2,900 kilometers (1,800 miles) thick and makes up about 84% of the Earth's total volume.
The crust, mantle, and core are the three coatings of the Earth. The crust is made up of minerals and solid rocks. The mantle lies under the crust and is largely made up of solid rocks and minerals, but it is punctured by pliable patches of semi-solid magma.
It is composed primarily of silicate rocks and is considered to be the source of the Earth's heat and most of its volcanic activity.
The mantle also plays a key role in plate tectonics, which is the theory that explains the movement of the Earth's continents and ocean floor.
Thus, the correct answer would be option (A) mantle.
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A rock is dropped from a cliff.
What will eventually happen to the rock's motion?
A. Gravity will cause the rock to slow down before it reaches Earth.
B. Air resistance will cause the rock to accelerate as it falls.
C. The rock will continue falling until Earth exerts a contact force on the rock.
D. Friction will cause the rock to continue in its state of motion.
Answer:
C
Explanation:
2057 Q.No. 9 a A typical car weighs about 1200 N. If the
coefficient of rolling friction is ur = 0.015. What horizontal
force is needed to make the car move with constant speed of
72 km/h on a level road? Also calculate the power developed
by the engine to maintain this speed.
Ans. 18 N, 360 W
Answer:
18N & 360W
Explanation:
Here,
Velocity of car (v) = 72 ∗ 1000 / 60∗60 = 20m/s
weight of car = 1200N
We have,
μr = Fs / R
Fs = μr ∗ R = 0.015 ∗ 1200 = 18N
Power developed by the engine = 18 ∗ 20 =360W
Hope this helps!!!what is the net force if you push a cart to the right with 5N of force, and a friend pushes the cart to the left with 5N of force?..
Answer: 0 newtons
Explanation: If you push a cart to the right with 5 newtons of force to the right and a friend pushes the cart 5 newtons to the left, there will be no motion.
To understand this further, take a look at the image I provided. I provided two arrows one of the left and on on the right. If two people are pushing on the cart with the same amount of force, the net force will be 0 so it will be a balanced force. When we have a balanced force, there's no motion.
a jet has a mass of 40000 kilograms the thrush pushing force of each of the four engines is 20000 Newtons what is the Jets acceleration when taking off
3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a hill. Describe the action and reaction forces that cause the locomotive to move up the hill, such as the reaction force and gravity.
This question has been posted before put the answers were not correct.
Answer:
Action - Pulling up the train.
Reaction - Friction on the locomotive
Explanation:
Locomotive is pulling the train upwards ,
Which is the action force applied by the locomotive,
As a reaction locomotive will be pulled by the train which is the reaction of pulling
Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.
For train action is pulling up by locomotive and reaction will be friction acting on it downwards.
As the skydiver falls to Earth, she experiences positive acceleration due to
Which of these is a landform created due to erosion by glaciers? Delta Horn Sandbars Spit
Answer:
The answer is option B, horn.
Explanation:
The landform horn was formed because when glaciers erode down to form a sharp peak which is called a horn. All of the other options are false because none of those land forms are formed because of the erosion of glaciers
Answer:
i think a delta
Explanation:
I might be wrong but I think it leaves behind a delta.
What is used to measure heat?
change in temperature
constant temperature
constant specific heat
change in specific heat
Answer:
change in temperature
Answer:
Change in temperature
Explanation:
did this on edge yesterday and that was the answer
If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s
The magnitude of the air drag is 784 N
Explanation:
An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:
[tex]F_D = mg[/tex]
where
[tex]F_D[/tex] is the magnitude of the air drag
m is the mass of the object
g is the acceleration of gravity
In this problem, we have
m = 80 kg is the mass of the airman
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting into the formula, we find:
[tex]F_D = (80)(9.8)=784 N[/tex]
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A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travels for a timet
consisting of only the car an open system or a closed system, and why?
Open system because the acceleration of the car is not constant.
Open system, because an external force is applied to the car that causes it to accelerate.
Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a timet
Closed system because mechanical energy was not removed from the system as a result of a net force
Rie
B
7:53 PM
12/11/2019
The car is considered as an open system, because an external force is applied to the car that causes constant acceleration of car. Hence, option (B) is correct.
The system isolated from its surrounding is known as closed-system. While, in an open system, the flow of information between the system and surrounding takes place, to cause any change in the system.
As per the concept of open and closed system:
The given problem indicates that car has initial acceleration of [tex]2 \;\rm m/s^{2}[/tex], travels for time t . Which means the car is accelerating constantly.And for a constant acceleration, there must be some external force acting on the car.Hence, the car must be considered an an open system, because the acceleration of car is related with the force, which is being applied from the outside of frame of reference.
Thus, we can conclude that considering only car is an open system because external force is applied to the car that causes it to accelerate. Hence, option (B) is correct.
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A person runs 5 km North and then turns west and travels for 7 km then turn south for 2 km it takes a person 4 hours to make the journey what is the persons distance, speed, displacement and velocity
1) Distance: 14 km
2) Speed: 3.5 km/h
3) Displacement: 7.6 km at [tex]23.2^{\circ}[/tex] north of west
4) Velocity: 1.9 km/h at [tex]23.2^{\circ}[/tex] north of west
Explanation:
1)
Distance is a scalar quantity, and it corresponds to the total length of the path covered by an object during its motion, regardless of its direction.
Here we have the following motion:
[tex]d_{y1}=5 km[/tex] North
[tex]d_x = 7 km[/tex] west
[tex]d_{y2}=2 km[/tex] south
So, the distance covered is:
[tex]d=d_x + d_y = 5 + 7 +2 = 14 km[/tex]
And since it is a scalar, distance has no direction.
2)
Speed is a scalar quantity telling "how fast" an object is moving. It is calculated as:
[tex]speed=\frac{d}{t}[/tex]
where:
d is the distance
t is the time taken
In this problem, we have
d = 14 km (distance)
t = 4 h (time taken)
Substituting,
[tex]speed=\frac{14 km}{4 h}=3.5 km/h[/tex]
And since it is a scalar, it has no direction.
3)
Displacement is a vector quantity connecting the initial position to the final position of motion. The magnitude is the short distance in a straight line between the initial and final position of motion.
Here we have the following motions:
[tex]d_{y1}=5 km[/tex] North
[tex]d_x = 7 km[/tex] west
[tex]d_{y2}=2 km[/tex] south
So the net displacement in the north-south direction is
[tex]d_y = 5 - 2 = 3 km[/tex] North
[tex]d_x, d_y[/tex] are at right angle, so they form the sides of a right triangle, of which the hypothenuse corresponds to the displacement. Therefore, we can find the magnitude of the displacement by using Pythagorean's theorem:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{7^2+3^2}=7.6 km[/tex]
And the direction is given by
[tex]\theta=tan^{-1}(\frac{d_y}{d_x})=tan^{-1}(\frac{3}{7})=23.2^{\circ}[/tex] north of west.
4)
Velocity is a vector quantity, given by:
[tex]velocity=\frac{d}{t}[/tex]
where
d is the displacement
t is the time taken
In this problem, we have
d = 7.6 km (displacement)
t = 4 h (time taken)
So, the velocity is
[tex]velocity=\frac{7.6 km}{4h}=1.9 km/h[/tex]
And being a vector, velocity has also a direction, which is the same of the displacement: so, [tex]23.2^{\circ}[/tex] north of west.
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Nichrome has a --------------melting piont
Answer:
Nichrome has __High__melting piont
Hope this helps
Answer:
high melting point
Nichrome is consistently silvery-grey in colour, is corrosion-resistant, and has a high melting point of about 1,400 °C (2,550 °F).
Explanation:
Allison needs to graph the yearly average snowfall in Alaska. She should use a:
A)bar graph
B)line graph
Answer: Line graph
Explanation:
Allison needs to graph the yearly average snowfall in Alaska. She should use a:
pie chart
pictograph
bar graph
line graph
You measure the voltage difference of a circuit to be 5 V and the resistance to be 1,000 ohms.
What is the current in the circuit?
Answer:
The current in the circuit will be [tex]5\times 10^{-3}\ Ampere[/tex].
Explanation:
Given:
Voltage difference = V = 5 V
Resistance = R = 1000 ohms
To Find:
Current, I = ?
Solution:
Ohm's Law:
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
i.e I ∝ V
∴ V = I × R
Where, I = current
V = voltage
R = resistance
Substituting the values of V and R we get
∴ 5 = I × 1000
[tex]I = \frac{5}{1000} \\I=0.005\ Ampere\\I=5\times 10^{-3}\ Ampere[/tex]
Therefore, the current in the circuit will be [tex]5\times 10^{-3}\ Ampere[/tex].
The current in the circuit can be found using Ohm's Law (I=V/R). The given values for voltage and resistance result in a current of 5 milliamperes.
Explanation:The subject matter in the question falls under the field of Physics, particularly Electricity. The problem is asking for the current in a circuit, based on given values for voltage and resistance. One can solve this by employing Ohm's Law, which describes the relationship between voltage (V), resistance (R), and current (I). Ohm's Law is represented by the formula I = V/R.
Substituting the given values into the formula, the current I is calculated as 5V / 1000Ω, which equals 0.005 Amperes, or 5 Milliamperes (mA). Thus, the current flowing in the circuit is 5 mA.
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Which of the following properties did Rutherford use in his experiment?
A. the negative charge of the alpha particles and the random distribution of protons
B. the negative charge of the alpha particles and the positive charge of the gold foil
C. the positive charge of the alpha particles and the negative charge of the electrons
D. the positive charge of the electrons in a uniform negative charge
Answer:
(C) The positive charge of the alpha particles and the negative charge of the electrons are the properties Rutherford used in his experiment
Explanation:
In his scattering experiment, scientist Ernest Rutherford used the property of positively charged alpha particles and negatively charge electrons. He performed this experiment by passing some of the alpha particles through a gold foil. The result was that the some alpha particles scattered while some passed through the gold foil without collision.
He concluded that the alpha particles are centrally positively charged and needed a large amount repelling force. This experiment of Rutherford is also known as Rutherford model of atom. This experiment helped him in doing so many other discoveries.
What distance will a car traveling 65km/hr travel in 3 hours
Answer:
it will cover 195km in 3 hours
Hey scooters dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. If the scooter is applying a force of 1850 N what is the coefficient of friction between a walk in the ground
The coefficient of friction is 0.363
Explanation:
There are two forces acting on the scooter in the horizontal direction:
- The applied force, F = 1850 N, forward
- The frictional force, [tex]F_f[/tex], backward
Since the scooter is moving at constant speed, the acceleration is zero, so the net force acting on the scooter must be zero. Therefore we can write:
[tex]F-F_f = 0\\F_f = F = 1850 N[/tex]
The frictional force can be written as
[tex]F_f = \mu R[/tex] (1)
where
[tex]\mu[/tex] is the coefficient of friction
R is the normal reaction of the ground on the scooter
For a flat horizontal surface, there is equilibrium along the vertical direction, so the normal reaction is equal to the weight:
[tex]R = W = mg[/tex]
where
m = 520 kg is the mass
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting into (1),
[tex]F_f = \mu mg = 1850 N[/tex]
and solving for [tex]\mu[/tex],
[tex]\mu=\frac{F_f}{mg}=\frac{1850}{(520)(9.8)}=0.363[/tex]
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Need help with 3 and 4
3) The boulder travelled for a distance of 10 m
4) The work done on the car is zero
Explanation:
3)
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem we have:
F = 24 N
W = 240 J
[tex]\theta=0^{\circ}[/tex], assuming that the force is applied in the same direction as the displacement
Therefore, we can find d, the displacement of the boulder:
[tex]d=\frac{W}{F cos \theta}=\frac{240}{(24)(cos 0)}=10 m[/tex]
4)
As in the previous exercise, the work done is
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
In this problem, we have:
F = 9000 N is the force applied
d = 0 is the displacement, since the car has not moved
Therefore, the work done on the car is
[tex]W=(9000)(0)(cos 0)=0[/tex]
So, no work is done on the car.
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In which situation is the object experiencing unbalanced forces?
Question 1 options:
A. A box resting on a horizontal floor
B. A car slowing as it reaches a stop light
C. A car with its cruise control set to 50 mph
D. A rocket sitting on the launch pad
B. A car slowing as it reaches a stop light
Explanation:
According to Newton's second law, the net force exerted on an object is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass of the object
a is its acceleration
This means that when an object is experiencing unbalanced forces ([tex]F\neq 0[/tex]), it must have a non-zero acceleration. Therefore, let's analyze the different options:
A. A box resting on a horizontal floor --> FALSE. Here the acceleration is zero, so no unbalanced forces.
B. A car slowing as it reaches a stop light --> TRUE. Since the car is slowing down, its velocity is changing, so the acceleration is non-zero and there are unbalanced forces.
C. A car with its cruise control set to 50 mph --> FALSE. Here the velocity is constant, so no acceleration and no unbalanced forces.
D. A rocket sitting on the launch pad --> FALSE. Here the rocket is stationary, so no acceleration, and no unbalanced forces.
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An object experiences unbalanced forces when the resultant force on it is non-zero. A car slowing as it reaches a stop light experiences unbalanced forces due to the brakes exerting a force, causing it to decelerate. Hence, the correct answer is B. A car slowing as it reaches a stop light.
In the question, we are asked to determine in which situation an object is experiencing unbalanced forces. An object experiences unbalanced forces when the resultant force acting on it is non-zero, leading to a change in its motion. If we analyze the options given:
A box resting on a horizontal floor is not experiencing unbalanced forces, because it is at rest and there are no net forces acting to change its state of motion (assuming no frictional losses).A car slowing as it reaches a stop light is experiencing unbalanced forces because the brakes are exerting a force opposite to the direction of motion, causing the car to decelerate.A car with its cruise control set to 50 mph is travelling at a constant velocity, indicating that the forces are balanced.A rocket sitting on the launch pad is not experiencing unbalanced forces if it is stationary and not in the process of takeoff.Therefore, the correct answer is B. A car slowing as it reaches a stop light, because this is when it is experiencing a net force that is causing a change in its motion.
Which statements apply to theories? Check all that apply.
A theory is a factual statement.
B theory is well-tested and widely accepted.
C theory is an explanation.
D theory cannot be revised or replaced.
E theory can help predict future events.
the answer is b,c, and e
Answer:
BCE
Explanation:
if you travel for three hours of a speed of 30 km/h, how far will you go?
Answer:
90km
Explanation:
every hour you travel 30km and you are traveling 3 hours 30×3
d) A wire carrying a current of 10A and 2m in length is placed in a field of flux density
0. 15T. Calculate the force on the wire if it is placed:
i) At 300
to the field
ii) Along the field (4 mks)
1) The force on the wire is 1.5 N
2) The force on the wire is zero
Explanation:
1)
The force experienced by a current-carrying wire in a magnetic field is given by
[tex]F=ILB sin \theta[/tex]
where
I is the current in the wire
L is the length of the wire
B is the magnetic flux density
[tex]\theta[/tex] is the angle between the direction of the wire and of the field
In this part of the problem we have
I = 10 A
L = 2 m
B = 0.15 T
[tex]\theta=30^{\circ}[/tex]
Substituting, we find
[tex]F=(10)(2)(0.15)(sin 30^{\circ})=1.5 N[/tex]
2)
As before, the force experienced by the wire is
[tex]F=ILB sin \theta[/tex]
where
I is the current in the wire
L is the length of the wire
B is the magnetic flux density
[tex]\theta[/tex] is the angle between the direction of the wire and of the field
In this case, however, the angle between the wire and the field is
[tex]\theta=0^{\circ}[/tex]
And therefore
[tex]sin 0^{\circ} = 0[/tex]
which means that the wire does not experience any force. In fact, the force experienced by the wire is zero when the wire is parallel to the field (such as in this case) and maximum when the wire is perpendicular to the field.
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Imagine a place in the cosmos..far from all gravitational and frictional influences..Suppose that you visit that place (just suppose)...and throw a rock.
What will the rock do? Why?
After the initial push, the rock will keep moving forever at constant velocity (constant speed in a straight line)
Explanation:
We can answer this question by using Newton's first law of motion:
"An object at rest (or in motion at constant velocity) will stay at rest (or will keep moving at constant velocity) unless acted upon unbalanced forces" (Law of inertia)
In this problem, we have a rock in a place very far from any force that can act on it. This means that there are no unbalanced force acting on it, so the rock will keep its state of motion forever.
In this situation, the rock is initially thrown by the astronaut. After the initial push, which accelerates the rock up to a certain velocity, there will be no more forces acting on the rock. This means that the rock will continue moving at a constant velocity forever, so at a constant speed in a straight line.
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