Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
Q)what are convex mirrors?
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
Why we use semiconductor instead of metal in thermopile.
Answer:
Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.
A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire
Answer:
the weight of the wire + 25kg
Explanation:
the rate of cooling determines ....... and ......
Answer:
freezing point and melting point
Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.
Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks
A=B=50NAngle=theta=35°
We know
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]
[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]
[tex]\\ \sf\longmapsto R=22.4i[/tex]
Resultant using the Law of Sines and the Law of Cosines will be R=95 N
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
The Magnitude of two forces =50 N
Angle between the forces = 35
By using the resultant formula
[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]
[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]
[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]
[tex]\rm R=\sqrt{5000+4500}[/tex]
[tex]\rm R=95\ N[/tex]
Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N
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A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.
Answer:
Florida is tropical, with a significant rainy seson
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?
you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth
Your friend would have a better experience of this event, than you .
What is an eclipse?An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,
there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.
For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.
The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.
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if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0?
Check attached photo
Check attached photo
Answer:
Explanation:
1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.
2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.
The magnitude of A and B must be equal
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity
The final speed of the ball when it reaches the floor is 7.10 m/s.
What is the conservation of energy?The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.
This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.
Here in the Question,
We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.
Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.
ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J
Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:
KEi + ΔPEi = KEf + ΔPEf
where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.
Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:
KEi = KEf + ΔKE
where ΔKE is the change in kinetic energy due to friction.
We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:
Wfric = ΔKE
where Wfric is the work done by friction.
The work done by friction can be expressed as:
Wfric = ffricd
where ffric is the force of friction and d is the distance traveled by the ball on the incline.
The force of friction can be expressed as:
ffric = μmg
where μ is the coefficient of kinetic friction, and m and g have their usual meanings.
Putting it all together, we get:
KEi = KEf + ffricd
KEi = KEf + μmgd
(1/2)mv^2 = (1/2)mu^2 + μmgd
v^2 = u^2 + 2gd
where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.
Plugging in the given values, we get:
v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)
v^2 = 50.405
v = 7.10 m/s
Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.
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Which circuit has the larger equivalent resistance: a circuit with two 10 ohm resistors connected in parallel or a circuit with two 10 ohm resistors connected in series?
Answer:
A circuit with two 10 ohm resistors connected in series.
Explanation:
The formula for the equivalent resistance for resistors in parallel is
[tex]\frac{1}{Rt} = \frac{1}{R1} + \frac{1}{R2}[/tex] So if R1=R2= 10 [tex]\frac{1}{Rt} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} <=> Rt =\frac{10}{2} =5 ohm[/tex]
The formula for the equivalent resistance for resistors in series is
Rt = R1 + R2 So Rt= 10 + 10 = 20
two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential
Answer:
B has greater potential
Explanation:
We know;
Potential Energy (PE) = mgh
where, m=mass of body
g=acceleration due to gravity
h=height of body
From the formula,
PE is directly proportional to the mass of the body
so the body with greater mass has greater potential.
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
