Hi can someon help me how to answer this?<br> Btw I'm from Philippines

A person is spinning an object around on a circular path on the end of a string of length 0.96 m. The object has a mass of 0.34

Rama09 [41]

I’m so sorry I needed points but I hope you get it right

8 0

3 years ago

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

Lana71 [14]

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the given Normal on block 2 = N

and $N = m_2 g \cos \theta$

and the tension in the given string is said to be $T = m_2 g \sin \theta$

When the acceleration $a=\frac{F}{m_1}$

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

$F = m_1g-T.$

It is known that if the said

$\theta=\frac{\pi}{2}$ ,

then Tension $T= m_2g$ $[since \sin(\pi/2) = 1]$ ,

Now making $"F = m_1g - m_2g"$

So If we are to make Force equal to zero

$F=0 => m_1g = m_2g \ or \ m_1 = m_2$

6 0

3 years ago

In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

4 years ago

0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc

weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

4 0

3 years ago

A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,

MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

$T - mg cos\theta = \frac{mv^2}{R}$

$T = mg cos\theta + \frac{mv^2}{R}$

now by energy conservation we have

$\frac{1}{2}mv^2 = mg(R cos\theta)$

$mv^2 = 2mgR cos\theta$

$T = mg cos\theta + 2mgcos\theta$

$T = 3mg cos\theta$

For string break down we have

$T = 2mg = 3mgcos\theta$

$cos\theta = \frac{2}{3}$

Now height from the ground is given as

$h = R - Rcos\theta$

$h = 24 - 24(\frac{2}{3})$

$h = 8 cm$

7 0

3 years ago

Hi can someon help me how to answer this? Btw I'm from Philippines