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2 years ago

For each of the problems below, you will need to draw a graph to find the solution.

Physics

1 answer:

Gelneren [198K]2 years ago

3 0

Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

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At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr

sattari [20]

Answer:bill 5 m/s. Jack:10 m/s

Explanation:

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2 years ago

1. Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomica

nydimaria [60]

Answer:

e telescopes

Explanation:

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2 years ago

Which best explains how an object at rest deep in space and far from any massive body behaves compared to an object in free fall

7nadin3 [17]

Answer: They behave the same because, according to the principle of equivalence, the laws of physics work the same in all frames of reference.

Explanation:

According to the equivalence principle postulated by Einstein's Theory of General Relativity, acceleration in space and gravity on Earth have the same effects on objects.

To understand it better, regarding to the equivalence principle, Einstein formulated the following:

A gravitational force and an acceleration in the opposite direction are equivalent, both have indistinguishable effects. Because the laws of physics must be accomplished in all frames of reference.

Hence, according to general relativity, gravitational force and acceleration in the opposite direction (an object in free fall, for example) have the same effect. This makes sense if we deal with gravity not as a mysterious atractive force but as a geometric effect of matter on spacetime that causes its deformation.

3 0

2 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

3 years ago

A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang

valina [46]

Answer:

work done is -2.8 × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

$W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta$

$W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta$

$W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta$

W = $-mgl[$ -cosθ $]^{0.047}_{0.045 }$

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8 × 10⁻⁶ J

Therefore, work done is -2.8 × 10⁻⁶ J

6 0

2 years ago