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iogann1982 [59]
3 years ago
12

A negative charge is moving across a room from south to north. A magnetic field runs from east to west. In what direction is the

magnetic force on the moving negative charge?
A.
toward the east
B.
toward the west
C.
toward the floor
D.
toward the ceiling
Physics
2 answers:
erica [24]3 years ago
6 0
Here is your answer

C. towards the floor

REASON:

Using Fleming's Left hand rule we can determine the direction of force applied on a moving charged particle placed in a magnetic field.

The direction of current will be just opposite to the direction of electron(negative charge) because current moves from positive to negative terminal whereas electron moves from negative to positive terminal.

So, direction of current- North to South

Now applying Fleming's Left hand rule we get the direction of force in downward direction, i.e. towards the floor.

HOPE IT IS USEFUL
elixir [45]3 years ago
5 0

B toward the West but east would be positive

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A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we
Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

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\Delta P = mgh

\Delta P = (8.7)(9.8)(3)

\Delta P = 255.78J

Therefore the change in the internal energy of the system is 255.78

7 0
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A weightlifter presses a 400 n weight 0.5 m over his head in 2 seconds, what is the power of the weight lifter
Scrat [10]
There are two ways to solve this problem. First we write the given.

Given: Force F = 400 N;  Height h = 0.5 m;  Time t = 2 s

Formula: P = W/t;  but Work W = Force x distance or W = f x d
 
Weight is also a Force, therefore:  W = mg, solve for Mass m = ?

m = w/g  m = 400 N/9.8 m/s²  m = 40.82 Kg

P = W/t = F x d/t  = mgh/t  P = (40.82 Kg)(9.8 m/s²)/2 s  

P = 100 J/s or 100 Watts

3 0
3 years ago
A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 seconds later?
Bogdan [553]
Given: 

V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
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For linear-motion problems with those given terms, the following formula is used:

V2 = V1 + as

Substituting the given values:

V2 = 24 + 2(8)
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Therefore the car will have a speed of 40 m/s as it leaves the tunnel.
4 0
3 years ago
A 7kg block slides down a 30 degree slope at a constant velocity of 2 m/s. How big is the force of friction acting on the block?
lisabon 2012 [21]

The force of friction is <u>34.3 N.</u>

A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity.  According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.

The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components  mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction   F acts upwards along the plane  and the normal reaction acts perpendicular to the plane.

Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.

F = mg sin\theta

Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.

F = mg sin\theta = (7 kg)(9.8 m/s^2)(sin30^o) =34.3 N

The force of friction is <u>34.3 N</u> up the plane.


6 0
3 years ago
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