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vladimir1956 [14]

3 years ago

15

A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3/2 km in 3.5 min. How fast (in m/s) is the car m

oving after this time?
(the equation that my teacher taught us is d= ((vi + vf)/2) x t (where d = distance, vi = initial velocity, vf = final velocity, t = time), so if you could use that it would be greatly appreciated.)

(please show all of your work or add a comprehensive explaination)

1 answer:

ArbitrLikvidat [17]3 years ago

7 0

Answer: no answer i dont no

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bonufazy [111]

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B) Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately 141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the ball is given by the equation for the range of a projectile as follows;

$Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}$

Substituting the known values, gives;

$Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m$

The horizontal range ≈ 141.25 m.

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Answer:

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What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the lef

Alex777 [14]

Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

Use the B-field formula:

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The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

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The current theory of the structure of the

irina1246 [14]

Answer:

pt 1: $m=1.66698*10^{21} kg$

Pt 2: $KE=1212.23531 J$

Explanation:

Information Given: (p = density)

l = 5200km d = 35km p = 2700kg/ $m^{2}$

Part 1: Mass

Find volume

$V=(l)^2(d)$
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$V=61.74*10^{16}$

Find Mass

$m=Vp$
$m=(61.74*10^{16})(2700)$
$m=1.66698*10^{21}$

Part 2: Kinetic Energy

$v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}$
$v=1.20497*10^{-9}$

$KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2$

$KE=1212.23531 J$

Part 3: Jogger Speed

set up, because I don't have the mass :(

Information given:

$KE_{jogger}$

$KE=\frac{1}{2}mv^2$
$v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }$

Input the values

Hope it helps :)

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3 years ago