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2 years ago

7

A distance-time graph indicates that an object travels 2m in 2s and then travels another 80m during the next 40s. what is the av

erage speed of the object

2 answers:

balandron [24]2 years ago

8 0

A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s. What is the average speed of the object?

a. 2 m/s

b. 4 m/s

c. 8 m/s

d. 10 m/s

whats the answer?

ipn [44]2 years ago

5 0

Answer:The average speed of the object is 1.95 m/s.

Explanation:

Distance traveled by an object in first interval, $T_1$ = 2 m

Time taken in firsts interval $T_1$ = 2 s

Distance traveled by an object in second interval interval, $T_2$ = 80 m m

Time taken in seconds interval $T_2$ = 40 s

$Average speed=\frac{total distance}{\text{Elapsed time}}=\frac{2 m+80 m}{2s+ 40 s}=1.95 m/s$

The average speed of the object is 1.95 m/s.

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

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3 years ago

Why do the other bulbs go dark when one bulb is removed in the series circuit but the other bulbs do not go dark when one bulb i

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Answer:

I think this

One bulb burning out in a series circuit breaks the circuit. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function.

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36. The top floor of a building is 20 m above the basement. Show that the water pressure in the basement is nearly 200 kPa great

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Explanation:

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When a spaceship is traveling at 99% of the speed of light (Lorentz factor = 7), an astronomer on board the ship will find thata

Natali5045456 [20]

Answer:

e) True. Measure the own values, so everything seems normal

Explanation:

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We call the time and the proper length the magnitude measured for an observer who does not move with respect to the measurement system.

In this case the astronomer is on the ship, for him he does not feel the movement of it, they are at rest with respect to each other. Therefore, their measurements are the so-called ones, this means that their values do not change since the two go at the same speed.

In examining the final statements we have

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