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Talja [164]
3 years ago
9

A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant planet. If 20 complete oscillation are

completed and 35.5 seconds, the acceleration is
a) 7.52 m/s squared

b) 9.81 m/s squared

c) 30.5 m/s squared

d) 42.0 m/s squared
Physics
1 answer:
givi [52]3 years ago
6 0

Answer :

Explanation :

It is given that,

length of pendulum, l = 0.6 m

Number of oscillation, n = 20

time, t = 35.5 seconds

so, Time period, T=\dfrac{t}{n}

T=\dfrac{35.5s}{20}

we know that the time period of pendulum is given by

T=2\pi\sqrt{\dfrac{l}{g}}

g=\dfrac{4\pi^2l}{T^2}

g=\dfrac{4\times(3.14)^2\times 0.6 \times (20)^2}{(35.5s)^2}

g=7.51\ m/s^2

or

g=7.52\ m/s^2

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Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
4 years ago
Which of the following statements is true of space exploration?
elena-s [515]
You don't have a following space exploration
6 0
3 years ago
A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 6 m/s in 5 seco
ddd [48]

Answer:

Explanation:

Acceleration =  a change in velocity / a change in time

Acceleration = ( final velocity - initial velocity) / a change in time

Acceleration = (6m/s - 35 m/s ) / 5 s

                     = (-29 m/s) /( 5 s)

                     = - 5.8 m/s^^2  

Remember Significant Figures  

- 6 m/s^2

P.S I have no idea why the answers say m/s because acceleration is m/s^2.

:)

7 0
3 years ago
Please answer i need help asap!!!
Alexus [3.1K]
The time to distance ratio is 2.1:1 , making the first time 5 seconds, the first distance 18.9 m, and the second time 15 seconds. I hope this helps!
5 0
4 years ago
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