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3 years ago

How to find final velocity

Physics

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

Explanation:

The equation or formula for velocity is similar to speed. To figure out velocity, you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

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The Zika Virus is most dangerous to which group of people?

laiz [17]

Answer:

option D pregnant women

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4 years ago

In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re

Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α = torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

6 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago

A dog, that has a mass of 16 kgs, runs across a yard. What is the average force applied to the dog from the ground as it runs ac

VikaD [51]

Answer:

156.96 N

Explanation:

F=ma where m is the mass and a is acceleration

Substituting 16 Kg for m and 9.81 m/s2 for g then

F=16*9.81= 156.96 N

4 0

3 years ago