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2 years ago

What is the length of a pendulum that has a period of 4. 89 seconds?

Physics

1 answer:

jeka942 years ago

6 0

Hi there!

We can use the following equation for a simple pendulum:

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

T = Period (4.89 s)
L = length (? m)
g = acceleration due to gravity (9.8 m/s²)

Rearrange the equation to solve for L.

$T^2 = 4\pi ^2 \frac{L}{g}\\\\L = \frac{gT^2}{4\pi ^2}$

$L = \frac{(9.8)(4.89^2)}{4\pi^2} = \boxed{5.936 \frac{m}{s}}$

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A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

4 0

2 years ago

When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq

Lilit [14]

Answer:60 gm

Explanation:

Given

initial velocity of ball $u=0$

Force exerted by racquet $F=540 N$

time period of force $t=5\ ms$

final velocity of ball $v=45\ m/s$

Racquet imparts an impulse to the ball which is given by

$J=F\Delta t=\Delta P$

$J=540\times \Delta t=m(45-0)$

$m=\frac{540\times 5\times 10^{-3}}{45}$

$m=60\ gm$

8 0

3 years ago

Who created the theory of general relativity?

pantera1 [17]

Who created the theory of general relativity?:

The answer would be: Albert Einstein.

Albert Einstein developed the general relativity theory (gravitation).

The year he developed the general relativity theory or (GR) was back in 1907 and 1915. Then many other contributions after 1915.

Thanks,

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7 0

3 years ago

Read 2 more answers

a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction 30° east of north

liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement $D_{1}$ = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey $D_{2}$ = 20 km (red arrow) is at an angle with reference to y-axis.