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Jet001 [13]
2 years ago
7

A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu

m must be shortened slightly to have a period of 1.00 s. What is true about the acceleration due to gravity on the distant planet?AnswerWe cannot tell because we do not know the mass of the pendulum.The gravitational acceleration on the planet is slightly less than g.The gravitational acceleration on the planet is slightly greater than g.The gravitational acceleration on the planet is equal to g.
Physics
1 answer:
Sergio [31]2 years ago
7 0

Answer:

The gravitational acceleration on the planet is slightly less than g.

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

The formula can also be rewritten as

g=(\frac{2\pi}{T})^2 L (1)

In this problem, we have a pendulum which has a period of T=1.00 s on Earth. The length of the same pendulum must be shortened on the distant planet to have the same period of T'=1.00 s: this means that the length of the pendulum on the distant planet, L', is shorter than the length of the pendulum on Earth, L

L'

By looking at formula (1), we see that g (the gravitational acceleration) is directly proportional to L. therefore, if L is shortened on the distant planet, it means that also the value of g is lower than on Earth:

so, the correct answer is

The gravitational acceleration on the planet is slightly less than g.

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Sedbober [7]

Answer:

Here's what I get  

Explanation:

A. Distance between A and B.

h = -½gt²

The stones go faster the farther they fall.

Stone A has already reached 5 m when B is released.

When B reaches 5 m, A has dropped further and is falling even faster.

The distance between the stones increases with time.

Figure 1 shows this effect in a graph of height vs. time.

B. Speed of Stone B

v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²  

v = 9.9 m/s

The stone is travelling at 9.9 m/s when it reaches 5 m.

C. Velocity vs time

v = -gt

Both stones accelerate at the same rate.

When Stone B has reached 10 m at time t, Stone A is falling much faster.

Fig. 2 shows this in a graph of velocity vs time.

 

3 0
2 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
What is the magnitude of the relative angle φ
melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0
2 years ago
A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward
Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

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                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

                                  ax=(change in velocity)/(Time)

                                  ax=\frac{-0.5-0}{3} =-\frac{1}{6} ms2

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2 years ago
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Studentka2010 [4]

Explanation:

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           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

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3 0
2 years ago
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