1. As a person stands on earth they put a force on the ground and the ground puts a force on them.

Describe some tests that could be done to determine what an unknown substance is.

natima [27]

Answer:

by finding melting and boiling points

Explanation:

3 0

3 years ago

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Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/

Pani-rosa [81]

Answer:

equation of motion for Bill is

$y(t) = 4.9t^2$

equation of motion for Ted is

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of motion for Bill is

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 0 + 0(t) +\frac{1}{2}gt^2$

$y(t) = \frac{1}{2}\times (9.8t)^2$

$y(t) = 4.9t^2$

equation of motion for Ted is

$y_0 = 2m -1m = 2m$

$y_0 = -4.2 m/s$

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2$

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

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3 years ago

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When can a car have no acceleration but still be moving

gregori [183]

Answer:

A. A car can still be moving without acceleration when it has some velocity.

B. A car can come to a stop when moving itself when it has some de acceleration.

C.A ca can start at 0 m/s and speed up to 15 m/s when it has some acceleration.

Explanation:

A. since the car has no acceleration the velocity of the car would help it moving.

B. when the car is moving with some velocity but coming to a stop this is possible of and only if it has some de acceleration

C. since the speed of the car increases from 0 to 15 m/s acceleration is involved. If the speed was constant there would be no acceleration involved.

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3 years ago

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A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe

ioda

You said 2 revolutions every 0.08 seconds

1 revolution = 2pi radians.

A). The 'unit rate' is (2 rev) x (2pi / 0.08 sec) = 50pi radians/sec. =

   157.1 radians per sec (rounded)

B). Radius of the wheel = 30 cm
   Circumference = 2pi R = 60pi cm = 188.5 cm (rounded)

   Rotation speed = 2 revs per 0.08 sec
   Linear speed = 2 x 60pi cm per 0.08 sec

   (120pi cm) / (0.08 sec) = 47.12 meters per sec

C). Frequency = (revs) per second

      = (2) / (0.08 sec) = 25 per second .

7 0

3 years ago

A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. de

uranmaximum [27]

Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a = the acceleration of descent., m/s².
Let u = initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u, v, and a are measured as positive upward.

Then
u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600 (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

3 0

3 years ago