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Arte-miy333 [17]

3 years ago

7

Two friends leave a movie theater and take different busses to the same ice cream shop. One bus takes a longer route driving on

a high-speed highway, while the other takes a shorter route on lower-speed local roads. Which friend has the greater displacement?

1 answer:

lisov135 [29]3 years ago

6 0

Answer: short displacement has shorter road and long displacement has longer displacement . now think your self Which one is right

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A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h

bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of 11 m for the first time.

b) Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

5 0

4 years ago

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

3 years ago

A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between

loris [4]

Answer:

d = 1.27m

Explanation:

Given m = 8.2×10-26kg, v = 1.1×10⁶m/s, q = +5e = 5×1.6×10‐¹⁹ C.

E = 49kN/C = 49000N/C

The displacement is given by

d = 1/2× mv²/qE = 1/2 × 8.2×10-²⁶ × (1.1×10⁶)²/(5×1.6×10-¹⁹ ×49000) = 1.27m

3 0

3 years ago

Read 2 more answers

What physical characteristic is the same for a particular substance regardless of the sample size .

jeka94

<span>The physical feel of H2O is the same regardless of the sample size of the substance. Water feels wet whether it is a single drop or equivalent in volume to an ocean. It is the way the substance feels no matter how much or how little there is present.</span>

4 0

4 years ago

Two magnets are sliding towards each other. One of the magnets, which has a mass of 125 grams, is moving in the positive x-direc

olasank [31]

Answer:

0.45 m/s in the negative x-direction

Explanation:

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

$m_1v_1+m_2v_2=(m_1+m_2)v_c$ where $v_c$ is the common velocity, $v_1$ and $v_2$ are velocities of magnet moving in positive x-direction and magnet moving in negative x-direction respectively, $m_1$ and $m_2$ are masses of magnet moving in positive x-direction and magnet moving in negative x-direction respectively.

Substituting 125 g for $m_1$ and 85 g for $m_2$ , 7.33 m/s $v_1$ , -11.9 m/s for $v_1$ then

$125 g\times 7.33 m/s + (85 g\times -11.9)=(125 g+ 85 g)\times v_c$

$-95.25 g. m/s=210 g v_c$

$v_c=\frac {-95.25 g.m/s}{210 g}=-0.453571429 m/s
\approx -0.45 m/s$

Therefore, the velocity of single unit is 0.45 m/s in the negative x-direction

5 0

3 years ago