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3 years ago

How is work related to force and displacement?

Physics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

Displacement is the distance and direction an object is moved. Force and work are directly proportional to each other, while force and displacement are indirectly propotional. The equation showing the relationship is W= Fd. OR Work is a scalar. The rate at which work is done or at which energy is transferred. Work is equal to the force parallel to the displacement.

Explanation:

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Suppose that a 710 kg car is traveling at 13 m/s. Its brakes can apply a force of 4000 N.

Maksim231197 [3]

Answer:lalalalalall

Explanation:

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3 years ago

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the truck trav

irina1246 [14]

Answer:

total time = 65 seconds

total distance = 1554 meters

Explanation:

kinematic equation:

final velocity = initial velocity + acceleration multiplied by time

v_1 = v_0 + at

28 m/s = 0 m/s + 2 m/s^2 (t)

t = 14 seconds

a) total time = 14 + 46 + 5 = 65 seconds

b) must solve for total distance and divide it by time.

d_1 = v_0t + 1/2 a * t^2

d_1 = 0 + 0.5(2) * 14^2

d_1 = 196 meters

d2 = vt

d2 = 28 *46

d2 = 1288 meters

v_1 = v_o + at

0 = 28 + a(5)

- 28/5 = a

a = - 5.6 m/s^2

d_3 = v_0t + 1/2 a * t^2

d_3 = 28 (5) - 0.5(5.6)*5^2

d_3 = 70 meters

total distance = d1 + d2 + d3 = 196 + 1288 + 70 = 1554 meters

4 0

3 years ago

What are the basic Sl units?

denis-greek [22]

The SI system, also called the metric system, is used around the world. There are seven basic units in the SI system: the meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd).

6 0

3 years ago

Read 2 more answers

Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3

Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

$F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N$

$F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}$

The net force on q3 is the sum of these two forces:

$F_{net} = F_{13} + F_{23} = -1.92\times 10^{-5} + (-19.5\times10^{-6}) = 2.8\times 10^{-5} N$

Since both forces are directed towards left, their sign should be negative.

5 0

3 years ago

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

4 years ago