Answer:
D. [tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Explanation:
The statement is not correctly written, the correct form is now described:
A particle with charge [tex]q = -1\,C[/tex] is moving in the positive z-direction at 5 meters per second. The magnetic field at its position is [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex]. What is the magnetic force on the particle?
From classic theory on Magnetism, we remember that the magnetic force exerted on a particle ([tex]\vec F_{B}[/tex]), measured in newtons, is determined by the following vectorial formula:
[tex]\vec F_{B} = q\cdot \vec v \,\times \,\vec B[/tex] (1)
Where:
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in teslas.
If we know that [tex]q = -1\,C[/tex], [tex]\vec v = 5\,\hat{k}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\vec B = 3\,\hat{i}-4\,\hat{j}\,\,\,[T][/tex], then the magnetic force on the particle is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\0\,\frac{C\cdot m}{s}&0\,\frac{C\cdot m}{s} &(-1\,C)\cdot (5\,\frac{m}{s} ) \\3\,T&-4\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = -(-4\,T)\cdot (-1\,C)\cdot \left(5\,\frac{m}{s} \right)\,\hat{i}+(-1\,C)\cdot\left(5\,\frac{m}{s} \right)\cdot (3\,T)\,\hat{j}[/tex]
[tex]\vec F_{B} = -20\,\hat{i}-15\,\hat{j}\,\,\,[N][/tex]
Which corresponds to option D.
What voltage is required to move 6A through 5Ω?
5. How does the existence of humans compare with Earth's age?
Answer:
Human activity has fundamentally changed our planet. We live on every continent and have directly affected at least 83% of the planet’s viable land surface. Our influence has impacted everything from the makeup of ecosystems to the geochemistry of Earth, from the atmosphere to the ocean. Many scientists define this time in the planet’s history by the scale of human influence, and label it as a new geological epoch called the Anthropocene.
Explanation:
A boat travels 28 m while it reduces its velocity from 27.5 m/s to 14.5m/s. What is the boat’s acceleration while it travels that distance?
Answer:
9.75 m/s²
Explanation:
Given that,
v= 14.5 m/s
u = 27.5 m/s
s = 28 m
a = ?
v² = u² -2as [ minus sign due to deceleration]
14.5² = 27.5² - 2 × a × 28
210.25 - 756.25 = -56a
-56a = -546
a = 9.75 m/s²
STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?
Staying with the dragster problem...two more... If the dragster did the amount of work you calculated in the previous problem in 4.38 seconds, how much power was generated in Watts?
Answer:
1,439,283.10Watts
Explanation:
The question looks incomplete. We need the workdone to get the power generated.
Power = Workdone/Time
Power = Force * Distance/Time
Let
Force = 24340 Newtons
Distance covered = 259meters
Given
Time = 4.38secs
Required
Power generated
Substitute the given parameters into the formula;
Power generated = 24340*259/4.38
Power generated = 6,304,060/4.38
Power generated = 1,439,283.10
Hence the amount of power generated is 1,439,283.10Watts
A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.
True
False
A marble is rolling off the edge of a table is observed to hit the floor 0.77 m from the table. If the top is 0.86 m high, how fast was the marble traveling when it left the table?
A.) 0.2 m/s
B.) 0.4 m/s
C.) 1.8 m/s
D.) 1.6 m/s
Answer:
B 0.4 have a nice day and hiiii
A 50 kg person steps off a diving platform that is 10 meters above the water below (Olympic height). With what speed do they hit the water?
Answer:
14 m/s
Explanation:
The following data were obtained from the question:
Mass = 50 kg
Initial velocity (u) = 0 m/s
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
The velocity (v) with which the person hit the water can be obtained as shown below:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 10)
v² = 0 + 196
v² = 196
Take the square root of both side
v = √196
v = 14 m/s
Therefore, he will hit the water with a speed of 14 m/s
Describe and give an example of mutualism.
Describe and give an example of commensalism.
Describe and give an example of parasitism.
Describe and give an example of competition.
Describe and give an example of predation.
Answer:
Mutualism - Bee to flower. Bee eats - flower reproduces
Commensalism - Tree Frog to plant or tree. Frog uses plant for protection.
Parasitism - Flea or tick to host. Parasite feeds off host.
Explanation:
Competition - relationship between organisms that strive for same resources. intraspecific and interspecific. ex) two males competing for mates.
predation - one organism kills and consumes another. wolf hunting moose, cat hunting mouse. venus fly trap killing insect
Ruby has an index of refraction of 1.7. Calculate the maximum angle with respect to the normal at which a light ray escapes a ruby if the ruby is in air.
a. 17°
b. 36°
c. 53°
d. 90°
e. Light escapes at all incident angles.
Answer:
Umm sorry just came for the points.
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
Answer:
a
[tex]\lambda = 3.68 *10^{-36} \ m[/tex]
b
[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 180 \ kg[/tex]
The speed of the person is [tex]v = 1 \ m/s[/tex]
The energy of the proton is [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]
Generally the de Broglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m * v }[/tex]
Here h is the Planck constant with the value
[tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]
=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]
Generally the energy of the proton is mathematically represented as
[tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]
Here [tex]m_p[/tex] is the mass of proton with value [tex]m_p = 1.67 *10^{-27} \ kg[/tex]
=> [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]
=> [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]
=> [tex]v = 3.09529 *10^{7} \ m/s[/tex]
So
[tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]
so [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]
=> [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Protists are unique organisms that are so different from each other that they are sometimes called the 'junk drawer' kingdom.
True or False
Answer:
true
Explanation:
Im in k12 and I got an 100%
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?
A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s
Answer:
B. 1275 kg*m/s
Explanation:
I = F(deltaT) = (deltaP) = mv2- mv1
Therefore,
I = mv2-mv1
m = 850 kg
v2 = 5 m/s
v1 = 3.5 m/s
I = (850)(5)-(850)(3.5)
I = 1275 kg* m/s
1
Betty is sitting on of her surfboard out in the ocean. She is waiting for the
perfect wave to come along so she can ride it in to shore. As she waits, she
notices that the waves roll by in patterns, or sets. As the top of each wave
passes by Betty, it pushes her up. Which part of the wave does this? *
Explanation:
yooooooooo
(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)
Answer:
Nope
Explanation:
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it around her head in a horizontal circle at 100 rpm. Assume the room temperature is 20 degrees Celsius. What are the highest and lowest frequencies heard by a student in the classroom?
Answer:Highest frequency =618.89Hz
Lowest frequency=582.22Hz
Explanation:
The linear velocity of a sound generator is related to angular velocity and is given as
Vs = rω where
r = the radius of circular path = 1.0 m
ω is the angular velocity of the sound generator. = 100 rpm
1 rev/min = 0.10472 rad/s
100rpm =10.472 rad/ s
Vs = rω
= 1m x 10.472rad/ s= 10.472m/s
A) Highest frequency heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,
f max = (v/ v-vs) fs
Where , v is the speed of the sound in air at 20 degrees celcius =
343 metres per second
vs is the linear velocity of the sound generator=10.472m/s
fs is the frequency of the sound generator= 600 Hz
f max = (343/ 343 - 10.472) x 600
=343/332.528) x600
=618.89Hz
B) Lowest frequency heard by a student in the classroom = Minimum frequency
f min = (v/ v+vs) fs
(343/ 343 + 10.472) x 600
=343/353.472) x 600
=582.22hz
Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?
Answer:
(a) the change in length of the silk is 0.001585 cm
(b) the maximum weight that a single thread can support is 17.67 N
Explanation:
Given;
mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg
length of the silk, L = 12 mm = 0.012 m
diameter of the silk, d = 0.15 mm
radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m
The cross sectional area of the silk;
A = πr² = π(0.075 x 10⁻³)²
A = 1.767 x 10⁻⁸ m²
The Young's modulus of elasticity of spider-silk is given by;
2.1 Gpa = 2.1 x 10⁹ N/m²
(a)
Apply Young's modulus of elasticity equation to determine the change in length of the silk;
[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]
[tex]x = 0.001585 \ cm[/tex]
(b)
the maximum weight that a single thread can support is given by;
[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]
The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²
[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]
if a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius
a 2m
ь 500m
C 5m
d.25m
Answer:
Explanation:
The centripetal acceleration is expressed as;
a = v²/r
a is the acceleration = 50m/s²
v is the velocity = 10m/s
r is the radius
To get the radius
r = v²/a
r = 10²/50
r = 100/50
r = 2m
Hence its radius is 2m
Sam heats an 8kg sample of sand, with a specific heat of 664 J/kg·C°, from 20° to 40°. What is the change in thermal energy?
Answer:
106.24 kJ.
Explanation:
Given that,
Mass of sample of sand, m = 8 kg
Specific heat of sand, c = 664 J/kg-°C
The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :
[tex]Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ[/tex]
So, the change in thermal energy is 106.24 kJ.
how does the uneaven heating of earths surface affects earths weather patterns
Answer: it causes some parts of the earth to get more radiation than others.
Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is present, given by B = 4.0i + 9.0x2 j , with x in meters and B in mT. Calculate the k-component of the force on the 2 m segment of the conductor that lies between x = 1.0 m and x = 3.0 m.
Answer:
0.546 [tex]\hat k[/tex]
Explanation:
From the given information:
The force on a given current-carrying conductor is:
[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]
where the length usually in negative (x) direction can be computed as
[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]
[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]
[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]
where;
current I = 7.0 A
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]
F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]
F = 0.546 [tex]\hat k[/tex]
At a young age, who has the greatest influence on your choices?
Why do stretching exercises increase flexibility more than cardio exercises?
Answer:
Explanation:
No da esta aplicación y si para que usted vella
which thermometer is used in hot region.why?
Answer:
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.
Explanation:
please mark me brainlist
Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.
How long does it take a vehicle to reach a velocity of 32 m/s if it accelerates from rest at a rate of 4.2 m/s^2?
What is the initial velocity of the vehicle?
What is the final velocity of the vehicle?
What is the acceleration of the vehicle?
Write the equation you will use to solve the problem.
How long does it take the vehicle to reach its final velocity?
0.13 seconds
18.1 seconds
7.62 seconds
134.4 seconds
Answer:
7.62
Explanation:
because you have to divide 32/4.2
and can you do a friend request so i can accept it
A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?
Answer:
(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰
(a)(ii) The maximum torque is 0.132 Nm
(b) The orientation of the coil is 45⁰
Explanation:
Given;
diameter of the circular wire, d = 8.6 cm = 0.086 m
radius of the wire, r = d /2 = 0.043 m
number of turns, N = 15 turns
magnetic field, B = 0.56 T
The torque on the wire is given by;
τ = NIABsinθ
where;
θ is the orientation of the wire
(a) maximum torque occurs when the orientation of the wire is at 90⁰
The maximum torque is given by;
τ = NIABsin(90⁰)
τ = NIAB
τ = (15)(2.7)(π x 0.043²)(0.56)
τ = 0.132 Nm
(b)
71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm
[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]
A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1 .What is the average acceleration of the ball?
Answer:
50m/s^2Explanation:
Step one:
given data
initial velocity u= 0m/s since the ball is at rest
time of contact t= 0.3s
final velocity v=15m/s
Required
acceleration a
from the first law of motion
v=u+at
substitute our given data
15=0+a*0.3
15=0.3a
divide both sides by 0.3
a=15/0.3
a=50m/s
The average acceleration is 50m/s^2
My favorite Naruto couple fighting together!!!! Remember this?! And remember.........this?!
I cried so hard when he died.
Answer:
ok tbh i didnt cry when neji died but now when i think about it i just cry thinking about how himawari could have met her uncle
Explanation:
A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. What is the vertical component of the ball’s velocity? What is the horizontal component of the ball’s velocity?
0 m/s; 70 m/s
61.8 m/s; 32.9 m/s
32.9 m/s; 61.8 m/s
70 m/s; 0 m/s
Answer:
answer is 61.8 m/s; 32.9
l am not sure
Using a maximum allowable shear stress of 70 MPa, find the shaft diameter needed to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 rev/min
Answer:
a
[tex]d = 0.0223 \ m[/tex]
b
[tex]d = 0.0481 \ m[/tex]
Explanation:
From the question we are told that
The maximum allowable shear stress is [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]
The power is [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]
considering question a
The shaft speed is given as [tex]v = 2500\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * P}{v}[/tex]
=> [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]
=> [tex]\tau = 152.8 \ N \cdot m[/tex]
Generally the maximum torque experienced by the shaft is mathematically represented as
[tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]
Generally diameter = 2 * radius (r)
So
[tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]
Generally the maximum allowable shear stress is mathematically represented as
[tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]
=> [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]
=> [tex]d = 0.0223 \ m[/tex]
considering question b
The shaft speed is given as [tex]v = 250\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]
=> [tex]\tau = 1528 \ N \cdot m[/tex]
Generally the shaft diameter is mathematically represented as
[tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]
=>[tex]d = 0.0481 \ m[/tex]