Which of the following tools measures weight? A balance A scale A cyclometer A graduated cylinder

Find the radius R of the orbit of a geosynchronous satellite that circles the earth. (Note that R is measured from the center of

laila [671]

Answer:

R^3 = GM / ω^2

R^3 = (6.67 * 10^-11) * (5.98 * 10^24) / (0.00007272)^2

R^3 = 7.54 * 10^22

R = 42,251,269

R = 4.225 * 10^7 m

4.225 * 10^7

Explanation:

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3 years ago

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A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$ (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

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3 years ago

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Determine the moment of the force at AAA about point PPP. Use a vector analysis and express the result in Cartesian vector form.

Dimas [21]

Answer:

τ = 0

Explanation:

At the moment it is defined

τ = F x r

In tete case they give us the strength and position in Cartesian form, so it is easier to solve the determinant

τ = $\left[\begin{array}{ccc}i&j&k\\F_{x}&F_{y}&F_{z}\\x&y&z\end{array}\right]$

Let's apply this expression to the exercise

a) P = (-6 i ^ -3j ^ -6 k ^) m

F = (-6 i ^ -3j ^ -6k ^) 103 N

τ = $\left[\begin{array}{ccc}i&j&k\\-6&-3&-6\\-6&-3&-6\end{array}\right]$

τ = i ^ (3 6 - 3 6) + j ^ (6 6 -6 6) + k ^ (6 3 - 3 6)

τ = 0

b) P = 24i ^ + 8j ^ + 9k ^

F = 24i + 8j + 9k

τ = i ^ (72-72) + j ^ (216-216) + k ^ (24 8 - 8 24)

τ = 0

c) P = -6i + 6j-4k

F = -6i + 6j-4k

τ = i ^ (24-24) + j ^ (- 24 + 24) + k ^ (-36 + 36)

τ = 0

.d) P = 24i-8j + 9k

Let's change the sign of strength

F = -24i + 8j-9k

Tae = (I j k 24 -8 9 -24 8 -9)

Tae = i ^ (72 -72) + j ^ (- 216 + 216) + k ^ (192-192)

Tae = 0

8 0

3 years ago

If a fox starts from rest and accelerates at a rate of 1.1 m/s^2, how long does it take the fox to run 5 m A. 3.0s B. 4.5s C. 9.

KATRIN_1 [288]

Given:
u=0 m/s
a=1.1 m/s^2
S=5 m
t=time it takes to run 5 m

Use the kinematics equation
S=ut+(1/2)at^2
=>
5=0*t+(1/2)1.1(t^2)
solve for t
t=sqrt(5*2/1.1)=3.015 seconds.

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4 years ago

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If a cheetah goes from 10 m/s to 30 m/s in 5 seconds, what was her acceleration?

grandymaker [24]

Explanation:

Y CHETOS XDXDXX

CDXCVHH

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3 years ago

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