First class lever and third class lever

When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second

vitfil [10]

F = ma
17 N = (3.75 m/s/s) m
m = 4.53 g

5 0

3 years ago

Read 2 more answers

(b) Figure 3.32 shows the orbit of a planet around the Sun. Compare the linear speed of the planet at positions X, Y and Z.

Viefleur [7K]

At Z ... slowest speed

At Y ... fastest speed

At X ... medium speed

Wherever it is in its orbit, the line from the planet to the Sun smears over the same amount of area every second.

That's Kepler's second law of planetary motion.

The reason this happens is: That's how gravity works. (A better explanation is available, but first you have to be able to twirl calculus and solid geometry in the air on long sticks.)

5 0

4 years ago

A racecar accelerates uniformly from 18.5 mil to 46.1 m/s in 2.47 seconds.

laiz [17]

Answer:

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

Explanation:

Uniformly Accelerated Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Following the definition above, the acceleration is defined as:

$\displaystyle a=\frac{v_f-v_o}{t}$

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:

$\displaystyle a=\frac{46.1-18.5}{2.47}$

$\displaystyle a=\frac{27.6}{2.47}$

$a = 11.17~m/s^2$

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

5 0

3 years ago

1.

Gwar [14]

Answer:

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

Explanation:

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...(1)

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of first object

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of second object

For 1:

We are Given:

$m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s$

Putting values in equation 1, we get:

$(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s$

Hence, the correct answer is Option C.

For 2:

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

$J=F\times t$

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

For 3:

Let the speed in right direction be positive and left direction be negative.

We are Given:

$m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s$

Putting values in equation 1, we get:

$(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s$

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

4 0

3 years ago

A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

zvonat [6]

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is 3.92 m/s.

Explanation:

Let's use Newton's 2nd Law to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

The mass "m" is the sum of the man and woman's masses: $85+65= 150 \ \text{kg}$ .

The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
$v=3.9158$

The Ferris wheel's tangential velocity is 3.92 m/s.

8 0

2 years ago