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A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spectrometer. What is the mass of this ion?

Answers

Answer 1

Answer:

Answer:

Mass of ion will be $22* 10^(-13)kg$

Explanation:

We have given ion is triply charged that is $q=3* 1.6* 10^(-19)=4.8* 10^(-19)C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7* 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=(mv)/(qB)$

$m=(rqB)/(v)=(0.36* 4.8* 10^(-19)* 7* 10^6)/(0.55)=22* 10^(-13)kg$

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Temperature°F = (9/5 * °C) + 32°°C = 5/9 * (°F - 32°) 1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:1. 5 cm = ________ mm2. 83 cm = ________ m3. 459 L = _______ ml4. .378 Kg = ______ g5. 45°F = ________ °C6. 80°C = _________ °F
A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90˚, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 × 10-3 m2, 50 turns, and a resistance of 180 Ω. During the time while it is rotating, a charge of 9.3 × 10-5 C flows in the coil. What is the magnitude of the magnetic field?

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?

C)What would the swimmer's drop time be if the bridge were twice as high?

Answers

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\nh=19.62 meters$

Part b

$v=u+at\nv=0+9.81*2\nv=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\nt^2=8\nt=2.83 secs$

A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.Calculate di/dt
di/dt = _________.

Answers

The value of di/dt from the given values of the solenoid electric field is;

di/dt = 7.415 A/s

We are given;

Number of turns; N = 450 per m

Radius; r = 1.17 cm = 0.0117 m

Electric Field; E = 8.2 × 10⁻⁶ V/m

Position of electric field; r' = 3.45 cm = 0.0345 m

According to Gauss's law of electric field;

∫| E*dl | = |-d∅/dt |

Now, ∅ = BA = μ₀niA

where;

n is number of turns

i is current

A is Area

μ₀ = 4π × 10⁻⁷ H/m

Thus;

E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)

Since we are looking for di/dt, then we have;

E(2πr') = (di/dt)(μ₀nA)

Making di/dt the subject of the formula gives;

di/dt = E(2πr')/(μ₀nA)

Plugging in the relevant values gives us;

di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)

di/dt = 7.415 A/s

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Answer:

$(di)/(dt) = 7.31 \ A/s$

Explanation:

From the question we are told that

The number of turns is $N = 450 \ turns$

The radius is $r = 1.17 \ cm = 0.0117 \ m$

The position from the center consider is x = 3.45 cm = 0.0345 m

The induced emf is $e = 8.20 *10^(-6) \ V/m$

Generally according to Gauss law

$\int\limits { e } \, dl = \mu_o * N * (di)/(dt ) * A$

=> $e * 2\pi x = \mu_o * N * (d i )/(dt ) * A$

Where A is the cross-sectional area of the solenoid which is mathematically represented as

$A = \pi r ^2$

=> $e * 2\pi x = \mu_o * N * (d i )/(dt ) * \pi r^2$

=> $(di)/(dt) = (2e * x )/(\mu_o * N * r^2)$ ggl;

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) \ N/A^2$

=> $(di)/(dt) = (2 * 8.20*10^(-6) * 0.0345 )/( 4\pi * 10^(-7) * 450 * (0.0117)^2)$

=> $(di)/(dt) = 7.31 \ A/s$

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Answers

Answer:13.6 cm

Explanation:

Given

v(image distance)=-8.5 m

height of object $(h_1)$ =6 mm

height of image $(h_2)$ =37.5 cm

and magnification of concave mirror is given by $m=(-v)/(u)=(-h_2)/(h_1)$

$m=(-37.5* 10)/(6)=-62.5$

$-62.5=(8.5* 100)/(u)$

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

$(1)/(f)=(1)/(v)+(1)/(u)$

$(1)/(f)=(-1)/(850)+(-1)/(13.6)$

$(1)/(f)=-0.00117-0.0735$

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

Final answer:

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

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A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.