Answer fast please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

snow_tiger [21]

Answer:

property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.

7 0

2 years ago

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An electron is moving directly toward you in a horizontal path when it suddenly enters a uniform magnetic field that is either v

blagie [28]

Answer:

To your left

Explanation:

The direction of the force exerted on charged particle due to a magnetic field is given by the right-hand-rule, where:

- The index finger indicates the direction of motion of the electron

- the middle finger gives the direction of the magnetic field

- the thumb gives the direction of the force if the particle is positively charged - otherwise, the direction is reversed

in this case, we have an electron (so, a negatively charged particle):

- The direction of motion (index finger) is horizontal, toward you

- The electron begins to curve upward as it enters the field, so this means that the force exerted on the electrons is upward --> the thumb must point downward (because the electron is negatively charged)

- The index finger gives us the direction of the magnetic field: therefore, to your left.

3 0

3 years ago

A spring attached to a mass is at rest in the initial

svet-max [94.6K]

Answer:

E=(1/2)kx^2

Explanation:

6 0

3 years ago

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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

2 years ago

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$