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3 years ago

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A ball is dropped from rest from the top of a building, which is 106 m high. The magnitude of the gravitational acceleration g =

9.8 m/s2 Keep 2 decimal places in all answers. In this problem, the following setup is convenient: Take the initial location of ball (the top of the building) as origin x0 = 0 Take DOWNWARD as +x (a) How far (in meters) does the ball fall in the first 3 s ?

1 answer:

Sergio039 [100]3 years ago

5 0

Answer:

44.1 m

Explanation:

initial velocity of ball, u = 0

height of building, H = 106 m

g = 9.8 m/s^2

t = 3 second

Let the ball travels a distance of h in first 3 seconds.

Use second equation of motion

$s=ut+\frac{1}{2}at^{2}$

h = 0 + 0.5 x 9.8 x 3 x 3

h = 44.1 m

Thus, the distance traveled by the ball in first 3 seconds is 44.1 m.

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Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107m above Earth's surface (that is, if it were three Earth ra

Vladimir79 [104]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
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3 years ago

A certain car's drive-train produces a force of 5300 N as it accelerates from 0

lakkis [162]

The power is $7.1\cdot 10^4 W$

Explanation:

First of all, we need to find the acceleration of the car, which is given by

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v = 60 mph = 26.8 m/s is the final velocity

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t = 10.0 s is the time

Substituting,

$a=\frac{26.8-0}{10.0}=2.68 m/s^2$

Now we can find the mass of the car by using Newton's second law:

$F=ma$

where

F = 5300 N is the force applied

m is the mass

$a=2.68 m/s^2$ is the acceleration

Solving for m,

$m=\frac{F}{a}=\frac{5300}{2.68}=1978 kg$

Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

And substituting,

$W=\frac{1}{2}(1978)(26.8)^2-0=7.10\cdot 10^5 J$

Finally, we can find the power output of the car:

$P=\frac{W}{t}$

where

W is the work

t = 10.0 s is the time elapsed

Substituting,

$P=\frac{7.1\cdot 10^5}{10.0}=7.1\cdot 10^4 W$

Learn more about power:

brainly.com/question/7956557

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4 years ago

What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

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We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

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3 years ago

When an object is at rest, it might have several forces acting on it. however,?

SSSSS [86.1K]

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3 years ago

If an airplane flies at 364 km/h, how long will it take to get from New York to Washington DC (288 km)? (If needed, round to the

Kazeer [188]

Answer:

0.791 h

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