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3 years ago

11

When the friction increases too much does the object stay still or moving opposite direction

1 answer:

goldfiish [28.3K]3 years ago

8 0

Answer:

the moment when the object starts to move

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Diffrence between FMRI and MRI?

love history [14]

FMRI creates the images or brain maps of brain functioning by setting up and utilizing an advanced MRI scanner in such a way that increased blood flow to the activated areas of the brain shows up on the MRI scan. The MRI scanners do not actually detect blood flow or other metabolic processes.

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3 years ago

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two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

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3 years ago

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The unit meters corresponds to what variable

Sati [7]

Do you have any options? My guess would be distance but I could be wrong.

7 0

3 years ago

The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me

kondaur [170]

Answer with Explanation:

The force of attraction between 2 charges of magnitude $q_1,q_2$ separated by a distance 'r' is given by

$F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}$

where

$\epsilon _o$ is a constant known as permitivity of free space

$k=9\times 10^{9}Nm^2/C^2$

Applying the given values in the above relation we get

$8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters$

5 0

3 years ago

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo

12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

$W = F\times d$

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute $W = 80.2\ J\ and\ d =25.0\ m$ in work done formula.

$80.2 = F\times 25$

$F=\frac{80.2}{25}$

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

3 0

3 years ago