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3 years ago

PLEASE HELP!!

Physics

1 answer:

Travka [436]3 years ago

7 0

Answer:

Walking while alternating knee lifts with each step.

Standing up and sitting down from a chair without using your hands.

Standing with your weight on one leg and raising the other leg to the side or behind you.

Explanation:

Those are the ones that I could best come up with.

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PLEASE HELP ME I NEED THIS ASAP!!!!

Zina [86]

Answer:

loud bangs

Explanation:

the pots for cooking fell

3 0

3 years ago

Read 2 more answers

A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0

3 years ago

Read 2 more answers

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

Margaret [11]

The vertical component is given by the sin of the angle. So this is 20*sin(53)=15.97m/s

8 0

3 years ago

A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to

Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

$\tau = I\alpha$

Where $\tau$ is the torque

$I$ is the moment of inertia

$\alpha$ is the angular acceleration

But, the angular acceleration is given by

$\alpha = \frac{\omega}{t}$

Where $\omega$ is the angular speed

and $t$ is time

Then, we can write that

$\tau = \frac{I\omega}{t}$

Hence,

$\omega = \frac{\tau t}{I}$

Now, to determine the angular speed, we first determine the Torque $\tau$ and the moment of inertia $I$ .

Here, The torque is given by,

$\tau = rF$

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ $\tau = 3.00 \times 195$

$\tau = 585$ Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

$I = \frac{1}{2}MR^{2}$

Where M is the mass and

R is the radius

∴ $I = \frac{1}{2} \times 325 \times (3.00)^{2}$

$I = 1462.5$ kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

$\omega = \frac{\tau t}{I}$

$\omega = \frac{585 \times 2.05}{1462.5}$

$\omega = 0.82$ rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0

3 years ago

A simple pendulum is made from a 0.54-m-long string and a small ball attached to its free end. The ball is pulled to one side th

Serga [27]

Answer:

0.37sec

Explanation:

Period of oscillation of a simple pendulum of length L is:

T = 2 π × √ (L /g)

L=length of string 0.54m

g=acceleration due to gravity

T-period

T = 2 x 3.14 x √[0.54/9.8]

T = 1.47sec

An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

The ball will first have V(max) at T/4,

=>V(max) = 1.47/4 = 0.37 sec

3 0

4 years ago