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ludmilkaskok [199]

3 years ago

13

Someone please help me for 7 points.... I don't know how to do this...

1 answer:

ololo11 [35]3 years ago

6 0

you need G, grav constant, here.

F=GMmoonMEllen/R^2

all other numbers given ... calculate ...

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Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

3 years ago

A body of mass 2kg moves round a circle of radius 5m with a constant speed of 10m/s. Calculate the force toward the centre of th

antiseptic1488 [7]

Answer:

F = 5

Explanation:

F = m x v^2/r = 2 x 10^2/5 =200/5 = 40 (N)

5 0

2 years ago

What is acceleration produced by a force of 12 newton exerted on an object of mass 3kg

jarptica [38.1K]

Answer:

a=F/m

a=12N/3kg (here newton can be written as kgm/s^2 so kg will be cancelled)

a=4m/s^2

Explanation:

3 0

3 years ago

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

galben [10]

Answer:

$\triangle P=1.95*10^{-4}$

Explanation:

Mass $m=0.001$

Diameter $d=1.2m$

Length $l=10m$

Generally the equation for Volume flow rate is mathematically given by

$Q=AV$

$V=\frac{Q}{\pi/4D^2}$

$V=\frac{0.001}{\pi/4(1.2)^2}$

$V=8.84*10^{-4}$

Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

$F=0.06$

Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

$H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}$

$H=19.9*10^{-9}$

Where

$H=\frac{\triangle P}{\rho g}$

$\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}$

$\triangle P=H*\rho g$

$\triangle P=1.95*10^{-4}$

6 0

3 years ago

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

kozerog [31]

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$

Solve for a;

$(V_2)^2=(V_1)^2+2ad$

Begin by subtracting $(V_1)^2$

$(V_2)^2-(V_1)^2=2ad$

Divide by 2d

$\frac{(V_2)^2-(V_1)^2}{2d} =a$

Now plug in your values:

$a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}$

$a=\frac{0-277.89m^2/s^2}{100m}$

$a=-2.78m/s$

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

$F=ma$

m = mass = 900kg

a = acceleration = -2.78m/s

$F=(900kg)(-2.78m/s)\\F=-2502N$

It's negative because the force has to be applied in the opposite direction that the car is moving.

8 0

3 years ago