In a region of space there is a uniform magnetic field pointing in the positive z direction. In what direction should a negative charge move to experience a force in the positive x direction

Answers

Answer 1

Answer:

the speed in -y

Explanation:

For this exercise we must use the right hand rule. The motion of a positive charge is given by.

Thumb points in the direction of speed

fingers extended in the direction of the magnetic field, + z axis

the palm in the direction of the force, as the charge is negative in the opposite direction of the force, axis + x

therefore the thumb is in the direction - y

the speed in -y


Related Questions



А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity

Answers

Thank lord for that please thank lord please thank

A car drives 110 km in 2 hours. Calculate the speed of the car

Answers

Answer: 55 kmph

Explanation: Divide 110 by 2

You’re working with a patient who suddenly falls. You should?

Answers

help them up and make sure there ok

If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N

Answers

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

What is Force?

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

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Convert 387.1 K to °C

Answers

387.1 kelvins = 113.95 degrees celsius

A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?

Answers

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical speed be right before it hits the ground?
A. 0 m/s
B. 15 m/s
C. 40 m/s
D. 30 m/s

Answers

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

Which of the following would likely happen if a person’s lactic acid system had difficulty breaking down glycogen in the muscles?
The person would have difficulty swimming across a lake.
The person would have difficulty sprinting in a race.
The person would have difficulty cycling down a hill.
The person would have difficulty running a marathon.

Answers

Answer:

The last one

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine

Answers

Fahrenheit scale is called the Rankine (°R) scale. These scales are related by the equations K = °C + 273.15, °R = °F + 459.67, and °R = 1.8 K. Zero in both the Kelvin and Rankine scales is at absolute zero.

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects

Answers

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Answers

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kgg = 9.8 m/s²T = ?

Use formula:

[tex]\boxed{\bold{T=m*g}}[/tex]

Replace and solve:

[tex]\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{T=98\ N}}}[/tex]

The tension in the rope is 98 Newtons.

Greetings.

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, what type of path will the particle follow

Answers

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

[tex]F = V\times B[/tex]

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright

Answers

It will become 100 times less bright

a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.

PLEASE HELP I NEED THIS TODAY

Answers

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

At what speed was object A moving ?

Answers

Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

What are impact and non-impact printers?​

Answers

Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.

In Non-Impact printers, no mechanical moving component is used.

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A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.

Answers

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

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A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter

Answers

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?

a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved

Answers

(B) linear momentum of the system is conserved

Which of these is a source of thermal energy inside earth

Answers

There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options

A body of mass 5 kg is moved by a horizontal force of 0.5 N on a smooth frictionless table for 20 seconds. Calculate the change in kinetic energy.

A. 5 J

B. 20 J

C. 10 J

D. 30 J​

Answers

Answer: 10 J

Explanation:

Trust ;)

is anyone online??just asking ​

Answers

Answer:

me...:(

Explanation:

Answer:

hello I'm online here thanks for the points (◔‿◔)

The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N

Answers

Answer:

8760 N

Explanation:

think this is the right answer :)

If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process. Express your answer in volts.

Answers

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.

It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex]    and   [tex]$\phi_{B,f} = 0$[/tex]

Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]

  [tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]

  = 30.27 V

Therefore, the emf generated is 30.27 V.

 

Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 48 s. What is Jason's top speed

Answers

Answer:

81.1 m/s

Explanation:

The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration

So, T - f = ma

T -  μmg  = ma

a = T/m - μg

susbstμituting the values of the varμiables into the equation, we have

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 2.67 m/s² - 0.98 m/s²

a = 1.69 m/s²

Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s

So, v = u + at

v = 0 m/s + 1.69 m/s² × 48 s

v = 0 m/s + 81.12 m/s

v = 81.12 m/s

v ≅ 81.1 m/s

So, Jason's top speed is 81.1 m/s

Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.

Answers

Answer:

4.408 [tex]\mathsf{M_{sun}}[/tex]

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]

combined mass (m₁+m₂)  = 4.408 [tex]\mathsf{M_{sun}}[/tex]

please help! will mark brainliest

Answers

Answer:

27.D.28.A

Explanation:

THE ANSWERME MY ANSWERWAH IS ME AND MY ANSWER WAHHHWALA NA SIYA WALA NA SIYA MASAYA MASAYA MASAYA WAHHtinapon na siya tipon na siya wahhsa basurahan wahh wahh ...............

kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay​

Answers

Explanation:

4. Alin sa mga sumusunod na awitin ang may tempong presto?

a. “Chua-ay”

c. “Akong Manok”

b. “Sitsiritsit

d. "Ili-ili Tulog Anay”

5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?

a. mabagal

c. mabilis na mabilis

b. mabilis at mabagal d. katamtamang bilis

6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?

a. descant

b. ostinato

c. tempo

d. texture

7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

TANONG:

Kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay[tex] ?[/tex]SAGOT:

Maayos, Maganda, at iba pa...

The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade

Answers

Answer:

The centripetal acceleration is 6.95 m/s²

Explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²

Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.

Answers

The answer is A because i know that it is and you will get the answer correct
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