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Gre4nikov [31]
3 years ago
7

To t

Physics
1 answer:
daser333 [38]3 years ago
4 0

Answer:

Welcome ShalomMHNCB Phx

Explanation:

the answer is because I just had this question and I got it wrong so te answer is 0.43

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What three variable factors determine the force of gravity between any two objects?
brilliants [131]
The force of gravity between two objects is:
F = G*m1*m2/r^2

So, it is dependent of the two masses and the distance between their centers of mass.  
5 0
3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
A horizontal compass is placed 21 cm due south from a straight vertical wire carrying a 36 a current downward. in what direction
Anit [1.1K]

 <span>
The needle of a compass will always lies along the magnetic field lines of the earth. 
A magnetic declination at a point on the earth’s surface equal to zero implies that 
the horizontal component of the earth’s magnetic field line at that specific point lies along 
the line of the north-south magnetic poles. </span>

The presence of a current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field. Since magnetic 
<span>fields are vector quantities, therefore the magnetic field of the earth and the magnetic field of the vertical wire must be combined vectorially. </span></span>

<span>
Where:</span>

B1 = magnetic field of the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T

B2 = magnetic field due to the straight vertical wire along the y-axis

We can calculate for B2 using Amperes Law:

B2 = μ₀ i / [ 2 π R ]

B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>

The angle can be calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>

θ = 53°

<span>
<span>The compass needle points along the direction of 53° west of north.</span></span>

8 0
3 years ago
Question 1 (1 point)
Dmitry [639]

1). trajectory

2). person sitting in a chair

3). 490 meters

4). 65 m/s

5). False.  The projectile's displacement, velocity, and acceleration have vertical and horizontal components, but the projectile doesn't.

6). False

7). The vertical component of a projectile doesn't change due to gravity, but the vertical components of its displacement, velocity, and acceleration do.

The vertical components do NOT equal the horizontal components.

8). Decreasing if you include the effects of air resistance.  Constant if you don't.  Gravity has no effect on horizontal velocity.

9). We can't see the simulation.  But if the projectile doesn't have jets on it, then as it travels upward, its vertical velocity must decrease, because gravity is trying to not let it get away.

10). We can't see the simulation.  But if the projectile is traveling downward, we would call that "falling", and its vertical velocity must increase, because gravity is pulling it downward.

6 0
3 years ago
Read 2 more answers
How long would it take for a person to sprint from the 0m line to the 100m line if they are traveling forward at 6m/s?
valina [46]

Answer:

  • <em><u>This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.</u></em>

Explanation:

<h3>Derive the kinematic equations for constant acceleration using integral calculus.</h3><h3>Use the integral formulation of the kinematic equations in analyzing motion.</h3><h3>Find the functional form of velocity versus time given the acceleration function.</h3><h3>Find the functional form of position versus time given the velocity function.</h3>
8 0
3 years ago
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