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2 years ago

PLEASE HELP I HAVE NO TIME PLEASEEE DONT SKIP

Physics

1 answer:

Triss [41]2 years ago

7 0

2m/s

it has to be 20 charecters just ignore this your answer is up there

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If the satellite has a mass of 3900 kg , a radius of 4.3 m , and the rockets each add a mass of 210 kg , what is the required st

ehidna [41]

Answer:

The required steady force of each rocket is 28.79 N

Explanation:

mass of the satellite, M=3900 kg

radius, r=4.3 m

mass of rocket, m=210 kg

time, t=5.4 min

Moment of Inertia:

I = 1/2 (Mr^2) + 4mr^2

I = 1/2 ( 3900* (4.3)^2) + 4 (210)*(4.3)^2

I = 51587.1 kg m^2

the angular acceleration is:

a= w/t

here w= 2*π*30

so,

a= 2*π*30 / 5.4* 3600

a=0.0096 rad/ s^2

the Torque becomes:

T=I*a = 4r*F

( 51587.1 )*(0.0096) = 4*4.3* F

F= 28.79 N

the required steady force of each rocket is 28.79 N

learn more about steady force here:

brainly.com/question/13841147

#SPJ4

7 0

1 year ago

Help 30 points and willgive brainliest

MissTica

HEJEGEJDHIEBDJDHDIDGDJGJDHD

8 0

2 years ago

Gravitational force is reduced by _____ between objects.

Fittoniya [83]

Gravitational force is reduced by:
B. The square of the distance..... hope that helps ;)

8 0

3 years ago

Read 2 more answers

An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

For m = 7 [g] and V = 2.3 [cm³]

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

For m = 10 [g] and V = 7 [cm³]

$Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$

For m = 15 [g] and V = 5 [cm³]

$Ro=15/5\\Ro=3[g/cm^{3} ]\\$

For m = 21 [g] and V = 8 [cm³]

$Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$

5 0

2 years ago

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

$R = u² sin2θ/g$

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

$u² = \frac{Rg}{sin2\theta}$

$u^2 = \frac{9.1 x 9.80}{sin26}$

$u^2 = 113.17$

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

$v^2 = (v_osin\theta)^2 -2gh$

$h =\frac{(v_o^2sin\theta)^2}{2g}$

$h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}$

h = 1.10 m

7 0

3 years ago