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Gwar [14]
3 years ago
13

If air resistance can be neglected, how does the acceleration of a ball that has been tossed straight upward compare with its ac

celeration if simply dropped?
Physics
1 answer:
timama [110]3 years ago
6 0
They are the same. If this is all happening on Earth, then the ball's acceleration is 9.8 m/s^2 in either case. That's the acceleration of gravity around here.
You might be interested in
g A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to th
abruzzese [7]

Answer:

L = 182.4 m      

Explanation:

Given:-

- The number of turns of the coil, N = 50

- The shape of the coil = square

- The angle between the coil and magnetic field, θ = 30°

- The change in magnetic field, ΔB = ( 700 - 250 ) μT

- The time duration in which magnetic field changes, Δt = 0.3 s

- The induced emf, E = 60.0 mV

Solution:-

- The problem at hand is an application of Faraday's law. The law states that the induced emf ( E ) is proportional to the negative rate of change of magnetic flux ( ΔФ / Δt ) and number of turns of the coil ( N ).

- The Faraday's law is mathematically expressed as:

                    E =  - N* ( ΔФ / Δt )

Where,

- The flux ( Ф ) through a current carrying with an cross-sectional area ( A ) at a normal angle ( θ ) to the direction of magnetic field ( B ) is given by the following relationship.

                    Ф = B*A*cos ( θ )

- We need the rate of change of magnetic flux ( ΔФ / Δt ) for the Faraday's law. I.e the induced emf ( E ) is proportional to rate of change in magnetic field ( ΔB / Δt ), rate of change of angle between the coil and magnetic field ( Δθ / Δt ) or rate of change of cross-sectional area of the coil under the influence of magnetic field.

- To determine the exact relationship. We will derive the multi-variable function of flux ( Ф ) with respect to time "t":

                     Ф ( B , A , θ ) = B*A*cos ( θ )

- The first derivative would be ( Use chain and product rules )

    ( ΔФ / Δt ) = ΔB / Δt*A*cos ( θ ) + B*ΔA/Δt*cos ( θ ) - B*A*sin ( θ )*Δθ/Δt

- For the given problem the only dependent parameter that is changing is magnetic field ( B ) with respect to time "t". Hence, ( ΔA/Δt = Δθ/Δt = 0 ):

                        ΔФ / Δt  = (ΔB/Δt)*A*cos ( θ )

- Substitute the rate of change of magnetic flux  ( ΔФ / Δt ) into the expression for Faraday's Law initially stated:

                        E =  - N*(ΔB/Δt)*A*cos ( θ )

- Plug in the values and evaluate the Area of the square coil:

                       A =  - E / ( N*(ΔB/Δt)*cos ( θ ) )

                       A = - 0.06 / ( 50*[ (250-700)*10^-6/0.3 ] *cos ( 30° ) )

                       A = - 0.06 / -0.07216

                       A = 0.8314 m^2

- The square coil has equal sides ( x ). The area of a square A is given by:

                      A = x^2

                      x = √0.8314

                      x = 0.912 m

 

- The perimeter length of a single coil in terms of side length "x" is given as:

                      P = 4x

Whereas for a coil of N turns the total length ( L ) would be:

                      L = N*P

                      L = 4Nx

                      L = 4 * 50 * 0.912

                      L = 182.4 m                 ... Answer

4 0
3 years ago
two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
Reika [66]

Answer:

t=750s

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

8 0
2 years ago
A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran
marshall27 [118]

Answer

a) For the rock

\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}

2sin\thetacos\theta = \dfrac{sin^2\theta}{2}

2cos\theta = \dfrac{sin\theta}{2}

tan\theta = 4

\theta = tan^{-1} 4

\theta = 76^0

b) \theta = 45^0 for maximum range

\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}

\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}

\dfrac{d_{max}}{d}=2.129

d_{max}=2.129 d

c) The value of θ is the same on every planet as g divides out.

5 0
3 years ago
An aurora occurs when _____.
drek231 [11]
The correct answer is A. Charged particles from the sun exite the atmosphere near the poles to create auroras.
6 0
3 years ago
Read 2 more answers
In a concave mirror, if an object is located between the center of curvature and the focal point, what is the best way to descri
alisha [4.7K]
Light rays travel parallel to each other, no image formed
7 0
3 years ago
Read 2 more answers
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