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bagirrra123 [75]
3 years ago
11

What must be known to determine the direction of the magnetic force on a charge? Check all that apply.

Physics
2 answers:
sergeinik [125]3 years ago
8 0
The strength of the magnetic field
Bingel [31]3 years ago
7 0
<h2>Answer:</h2><h3>A, C, D</h3><h2>Explanation:</h2>

What must be known to determine the direction of the magnetic force on a charge? Check all that apply.

<h3>A- the type of the charge </h3>

B- the amount of the charge

<h3>C- the direction of the magnetic field </h3><h3>D-  the velocity of the charge</h3>

E- the strength of the magnetic field

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Wind blowing sand from one location to another is an example of
Sedaia [141]
Its an example of weathering
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3 years ago
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A coin is dropped in a 15.0 m deep well.
labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0
3 years ago
A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is r
lawyer [7]

Answer:

New time period, T_2=2.12\ s

Explanation:

Given that,

Mass of the object 1, m_1=0.75\ kg

Time period, T_1=1.5\ s

If object 1 is replaced by object 2, m_2=1.5\ kg

Let T_2 is the new period of oscillation.

The time period of oscillation of mass 1 is given by :

T_1=2\pi \sqrt{\dfrac{m_1}{k}}

1.5=2\pi \sqrt{\dfrac{0.75}{k}}............(1)

The time period of oscillation of mass 2 is given by :

T_2=2\pi \sqrt{\dfrac{m_2}{k}}

T_2=2\pi \sqrt{\dfrac{1.5}{k}}............(2)

From equation (1) and (2) we get :

(\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}

(\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}

\dfrac{1.5}{T_2}=0.707

T_2=2.12\ s

So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.

3 0
3 years ago
If an object is accelerating at a rate of 25 m/s2 how fast will it be moving in 1.50 min
AURORKA [14]

1.5 min = 90 sec

(25 m/s²) x (90 sec) =

(25 x 90) (m - s / s²) = 2,250 m/s .

At the end of 1.5 min of acceleration, the object will be moving 2,250 m/s faster than it was at the beginning of the 1.5 min.

We don't know how fast that is, because you never told us how fast it was moving when the acceleration started.

(I'd stand out of the way of that object if I were you.  2,250 m/s is a little over 5,000 miles per hour !  And that's on top of whatever speed it had before it started accelerating ... at better than 2.5 G's !)

7 0
3 years ago
Read 2 more answers
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
3 years ago
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