Ask question

Ask question

All categories

jek_recluse [69]

2 years ago

10

Un cuerpo se desplaza del punto B al punto A. Si la posición del punto A es 77 m y la del punto B es -51 m. El desplazamiento de

l cuerpo es

1 answer:

Sliva [168]2 years ago

8 0

Answer:

bssnsjsbsjsjdhsjs gsjs

Explanation:

hhshdhd sndjsj dbdjd

You might be interested in

Centripetal is a word used in science to mean

bogdanovich [222]

: the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation a string on the end of which a stone is whirled about exerts centripetal<span> force on the stone — compare centrifugal force.</span>

7 0

3 years ago

A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15

cupoosta [38]

Answer:

$\boxed {\boxed {\sf 4.5 \ m/s \ in \ the \ positive \ direction}}$

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

$v_f= v_i + at$

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

$v_i$ = 2.7 m/s
a= 0.15 m/s²
t= 12 s

Substitute the values into the formula.

$v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)$

Multiply the numbers in parentheses.

$v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)$

$v_f = 2.7 \ m/s + (0.15 \ m/s *12)$

$\v_f=2.7 \ m/s + (1.8 \ m/s)$ $v_f=2.7 \ m/s + (1.8 \ m/s)$

Add.

$v_f=4.5 \ m/s$

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0

2 years ago

Can someone help me on question 1?

grigory [225]

In question 1, both of your answers are correct, but I don't understand the process you went through in the 'a' part.

R = v/I . That's a correct formula.
But it doesn't help you in this form, because you need to find I
So turn it into a helpful form ... Solve it for I, so it says I=something.

R= v/I

Multiply each side by I : R I = V.

Now divide each side by R: I= V/R .
THERE'S the equation you want.

I = V / R

I = 1.5 / 10 = 0.15 Amp.

That's slightly cleaner, although I don't really understand what you were actually thinking in that part.

But again ... You answered both parts correctly, and your process in b is fine.

6 0

3 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

3 years ago

Simplicity expression

Volgvan

They have expression<span>, modesty, sex appeal, </span><span>simplicity.</span>

7 0

3 years ago