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3 years ago

W imieniu państwa napisz zaproszenie dla rodzeństwa na uroczystą kolacje

Physics

1 answer:

HACTEHA [7]3 years ago

8 0

On behalf of the state, write an invitation for the siblings for the gala dinner

Hello there, we would like to welcome you to the gala dinner hope you can come :)

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A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.

gulaghasi [49]

The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force $F_f$ acting against the motion. Since their resultant must be zero, we have:
$F-F_f = 0$
The frictional force is
$F_f = \mu mg$
where
$\mu=0.4$ is the coefficient of kinetic friction
$mg=400 N$ is the weight of the block.

Substituting these values, we find the magnitude of the force F:
$F=F_f = \mu mg=(0.4 )(400 N)=160 N$

4 0

3 years ago

Which is an example of a solution?

vladimir1956 [14]

Answer:

Explanation:

plato

6 0

3 years ago

A motorboat accelerates uniformly from a velocity of 10.5 m/s to the west to a velocity of 6.5 m/s to the west. If its accelerat

BlackZzzverrR [31]

Answer:

7.4m ............................

3 0

3 years ago

If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

7 0

3 years ago

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$

6 0

3 years ago