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3 years ago

11

a charge feels a 2.89*10

lign="absmiddle" class="latex-formula"> N force when it moves 288m/s perpendicular (90) deg to a magnetic field of $2.77*10^-5$ T. how big is the charge?

1 answer:

mojhsa [17]3 years ago

8 0

3.62, -5
Explanation: I got it correct on a acellus

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As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length

grin007 [14]

Answer:

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

Question:

If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by

vb=Lwt+gt2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

The correct answer is

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5 = vb² +2g×vbt+1/2gt²

Explanation:

We have from the relation

v = u + gt, S = ut + 1/2 gt², v² = u² + 2gS

in this case S = hb, u = vb=Lwt+gt2. and v = vground

therefore v² = (Lwt+gt²)² + 2 × g × hb

= (Lwt+gt²)² + 2 × g × (Lwt+gt²)×t + 1/2 gt² = vb² +2g×vbt+1/2gt²

vground = (t²+lw²+(2t³+2t²)×g×Lw+(t⁴+2t³+t²) ) ∧ 0.5

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3 years ago

What is the relationship between matter and energy as it changes states of matter (phase changes?)

AlekseyPX

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.

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4 years ago

A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1

Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

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3 years ago

HELP 10 POINTS <br> Which set of balloons would exhibit a greater electric force and why?

vovikov84 [41]

Answer:

The second one because there is less <em><u>distance between them.</u></em>

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3 years ago

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou

Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where $\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$

$v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}$

$v_{hp}=\sqrt{\frac{2mgh}{m+m)}$

$v_{hp}=\sqrt{gh}$

For solid cylinder:

$I_{sc}=\frac{1}{2}mR^2$

$v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}$

$v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}$

$v_{sc}=\sqrt{\frac{3}{4}gh}$

For solid sphere,

$I_{ss}=\frac{2}{5}mR^2$

$v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}$

$v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}$

$v_{ss}=\sqrt{\frac{10}{7}gh}$

Then comparing the speed of the three objects we have:

$v_{hp}$

$\sqrt{gh}$

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4 years ago