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3 years ago

6

E) the time it takes the car to reach 20.0ms-1

1 answer:

Galina-37 [17]3 years ago

8 0

4.2 thats the answer

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The following coplanar forces pull on a ring 200N at 30 degrees and 500N at 80 degrees and 300N at 240 degrees and an unknown fo

Alex73 [517]

MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)

<h3>What is the formula of magnitude?</h3>

The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)

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3 0

2 years ago

What are the negative effects of retirement?

yarga [219]

Answer:

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5 0

3 years ago

Read 2 more answers

a 0.0818 kg salt shaker on a rotating table feels an inward frictional force of 0.108 N when it is moving 0.333m/s. what is the

serg [7]

<u>Answer:</u>

<em>The required radius of its motion is $0.084m$ .</em>

<u>Explanation:</u>

The formula for calculating the required radius of its motion is given by

$F = (mv^2)/r$

Where <em>m= mass </em>

<em>V= moving velocity </em>

<em>F=frictional force </em>

<em>r = radius of its motion </em>

Then the required radius of its motion is given by

$r =(mv^2)/F$

<u>Given that </u>

<em>mas =0.0818 kg </em>

<em>Frictional force= 0.108 N </em>

<em>Moving with Velocity of = 0.333 m/s </em>

<em>radius of its motion = $\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}$ </em>

<em>Hence the required radius of its motion is r = $8.4 cm=0.084m$ </em>

6 0

3 years ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

5 0

3 years ago

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a v

Umnica [9.8K]

Answer:

Average force exerted by the water on the blades is 1024 N

Explanation:

As we know that the force due to water jet is given by

$F = (v_{in} - v_{out})\frac{dm}{dt}$

here we know that

$v_{in} = 16 m/s$

$v_{out} = -16 m/s$

$\frac{dm}{dt} = 32 kg/s$

so here we will have

$F = (16 + 16)(32)$

$F = 1024 N$

Average force exerted by the water on the blades is 1024 N

6 0

3 years ago