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4 years ago

9

A figure skater glides along a circular path of radius 3.42 m. If she coasts around one half of the circle, find the magnitude o

f the displacement vector.

1 answer:

dusya [7]4 years ago

3 0

Displacement is the distance between the start- and end-points,
no matter what route is followed from start to finish.

If she travels around half of the circle, then her start- and end-points
are the two ends of a diameter, and her displacement is its length.

The length of the diameter is (2 x radius) = 6.84 meters.

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

$P=2.62atm$

Thus, the final pressure in the two containers is, 2.62 atm

8 0

3 years ago

The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......

valina [46]

Answer:

s^ -1 ( or 1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec or m/sec

so b would have units of s^-1

(or perhaps a more general term would be time^-1)

8 0

2 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

What is the length of a simple pendulum with a period of 2.0 s?

Mariana [72]

we know

T=2.pie sqr root(l/g)

l= (sqrof 7/22)*9.8

L= .99 or approx 1m

7 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

$v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}$

$v_{y}=u Sin 30 - 2u Sin 30$

vy = - u Sin 30

The magnitude of final velocity is given by

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}$

v = u

Thus, the velocity is same as it just reaches the ground.

6 0

4 years ago