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erica [24]
3 years ago
9

WHEN THE FORCE APPLIED ON AN ELASTIC OBJECT IS EQUAL AND OPPOSITE,THEN THIS FORCE MAY

Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

Elastic force

Explanation:

Elastic force occur when deformed or elastic object try to return to its original or natural length. When an elastic material is stretched, it resist it change in shape. The force act to return the elastic material to its original shape, it's then exert a counter force in opposit e direction.

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Please show your workmkmmkekdmnesnndndndndn
jekas [21]

Answer:

Explanation:

Remark

The question really is how long does the projectile stay in the air if it was just dropped from a height of 46 meters.

Givens

vi = 0

a = 9.81 m/s^2

h = 46 m

t = ?

Solution

d = vi*t + 1/2 a t^2

46 = 0 + 1/2 * 9.81 * t^2       multiply both sides by 2

92 = 9.81 * t^2                     divide by 9.81

t^2 = 9.38                            Take the square root of both sides

sqrt(t^2) = sqrt(9.38)

t = 3.06 seconds.

In 3.06 seconds the object will hit the ground, thus stopping it. How far horizontally does it go in that time.

d = r * t

d=?

r = 102 m/s

t = 3.06 seconds

d = 102 * 3.06

d = 312.4  meters.

4 0
3 years ago
Which body part is included in circulatory system
velikii [3]
The heart and the blood vessels.
6 0
4 years ago
A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position
Illusion [34]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

X(t)=\int V_{x}(t)dt

       =\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C

\to t=0\\\\ \to X(0)=2.3 \ m

\to X(0)=0+C\\\\ \to C=2.3\  m

\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\

        = \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\  = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m

In point b:

when t=2.2 \ s

the Particle velocity  (V)=2 \times 2.22 =9.68\  \frac{m}{s}

In point c:

Calculating the Particle acceleration:

\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2  =8.8 \ \frac{m}{s^2}

8 0
3 years ago
Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys
xxMikexx [17]

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

3 0
3 years ago
Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a
Marianna [84]

Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

4 0
3 years ago
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