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3 years ago

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Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat

ion of aB = -11 m/s2. The cars are initially at rest and separated by a distance d = 2900 m on the x-axis. In how many seconds does Car B reaches to Car A?

1 answer:

Paladinen [302]3 years ago

4 0

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² = 2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

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A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

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Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

3 years ago

An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0

sleet_krkn [62]

Answer:

22.26 years

, 15.585 light years , 11.13 light years

Explanation:

a)

$t' = t/(\sqrt{1-(v/(c*v)/c)}$

= $15.9/\sqrt{(1-0.7*0.7)}$

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years

c)

0.7*c*15.9

=11.13 light years

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3 years ago

A meter stick hurtles through space at a speed of 0. 95c relative to you, with its length perpendicular to the direction of moti

marta [7]

Answer:

Explanation:

Caty , Use the relativity formula for length. ( they teach this in H.S. ? ) it's from my Modern Physics in college, A 300 level class

L = $L_{0}$ $\sqrt{1-\frac{v^{2} }{c^{2} } }$

L = 3 $\sqrt{1-\frac{.95^{2} }{1^{2} } }$

L = 0.9367496998 meters

L = 0.94 meters approx

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2 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

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3 years ago

A student throws a water balloon with speed v0 from a height h = 1.82 m at an angle θ = 29° above the horizontal toward a target

3241004551 [841]

Answer:

$v_o = 7.76 m/s$

Explanation:

As we projected the balloon at speed vo at an angle of 29 degree

so the two component of velocity is given as

$v_x = v_ocos29 = 0.875 v_o$

$v_y = v_o sin29 = 0.485 v_o$

now we know that in x direction we have

$d = v_x t$

$7.5 = 0.875 v_o t$

$v_o t = 8.57$

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$- 1.82 = (0.485 v_o) t - \frac{1}{2}gt^2$

$-1.82 = 0.485(8.57) - 4.9 t^2$

$t = 1.1 s$

now we have

$v_o = 7.76 m/s$

7 0

3 years ago