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Sladkaya [172]
2 years ago
7

A stone of mass 5g is projected with a rubber catapult. if the catapult is stretched through a distance of 7cm by an average for

ce of 70 North, calculate the instantaneous velocity of the stone when released ​
Physics
1 answer:
weqwewe [10]2 years ago
4 0

Explanation:

solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...

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Physics. Need help. Brainlieast answer for most/ all of the answers answered
Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit;  its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both  hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds;   b). 13.1 meters;   c). 17.6 m/s down;   d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s;  b). 29.7 m/s

10). 7.24 meters

11).  700 meters

12).  A). 103.7 meters ( ! she's in big trouble ! );     B).  17.5 meters

3 0
3 years ago
We have discovered some exoplanets that are still forming from a nebula. How might those planets change over time?
Rudik [331]
Based on several theories made by scientists, planets are formed because of the accumulation of gases and other particles that are attracted to each other. These accumulated gases form into clumps and eventually the clumps get bigger and turn into a big orbital mass. The exoplanets may experience change over time through the observance of its orbit in a particular axis, and if there are other debris that might affect the planet's continuous growth.
6 0
3 years ago
What change will always result in an increase in gravitational force between two objects
tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.


Since the formula is


F = G\frac{m_1m_2}{d^2}


If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.


So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.


So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0
4 years ago
A scientist is studying a shock wave from an earthquake. What kind of wave is being studying?
Pavel [41]

Answer:

Longitudinal Mechanical Wave

Explanation:

Mechanical waves are the waves that require medium to propagate. And a longitudinal wave is a wave in which the vibration of the energy(here: mass specifically) is in the direction of propagation of wave.

Shock wave, strong pressure wave in any elastic medium such as air, water, or a solid substance, produced by supersonic aircraft, explosions, lightning, or other phenomena that create violent changes in pressure.

Shock waves travel faster than sound and their speed increases as the amplitude of the wave is increased but their intensity fades faster due to the fact that some of its energy gets expended in the form of heat due to the resistance of the medium.

3 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
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