SYW A force of 175 N is needed to keep a stationary engine of weight 640 N on wooden skids from

How are Earth and Venus similar? How is Venus different from Earth? (Provide two similarities and two differences.)

Alika [10]

Venus shares a similar size, surface composition, and has an atmosphere with a complex weather system. Venus is different from Earth because it spins the opposite direction of Earth and it’s rotation is very slow.

8 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

3 years ago

A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.

elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

6 0

4 years ago

Read 2 more answers

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
This friction force has a maximum value, that can be written as follows:

$F_{frmax} = \mu_{s} *F_{n} (1)$

where μs is the coefficient of static friction, and Fn is the normal force,

perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$