The work that a force does by acting on an object is equal to what?

1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?

VARVARA [1.3K]

Answer:

2km

Explanation:

Given data

We are told that the direction traveled are

North>>>East>>>South

Hence the displacement is defined as the distance away from the initial position is

Initial position =18km

FInal position = 16km

The displacement = 18-16= 2km

Hence the displacement is 2km

6 0

3 years ago

2. Willingness to take turns is one way we can express our attitudes through

leva [86]

B because willingness shows confidence in your answer

3 0

3 years ago

Q: What happens when cold air approaches a body of warm air? Optional Answers: A. The Warm air rises. B. The warm a

artcher [175]

A. The warm air rises! Hope this helps

8 0

3 years ago

Read 2 more answers

A cube has a density of 1900 kg/m 3 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experime

Deffense [45]

Answer:

4847.94844926 kg/m³

Explanation:

$\rho'$ = Actual density of cube = 1900 kg/m³

$\rho$ = Density change due to motion

v = Velocity of cube = 0.92c

c = Speed of light = $3\times 10^8\ m/s$

Relativistic density is given by

$\rho=\dfrac{\rho'}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-\dfrac{0.92^2c^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-0.92^2}}\\\Rightarrow \rho=4847.94844926\ kg/m^3$

The cube's density as measured by an experimenter in the laboratory is 4847.94844926 kg/m³

3 0

3 years ago

A 0.032-kg bullet is fired vertically at 230 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined b

gogolik [260]

Answer:

$heigth=83.44m$

Explanation:

Given data

Baseball mass m₁=0.15 kg

initial speed v₁=0

Bullet mass m₂=0.032 kg

final speed v₂=230 m/s

To find

height h=?

Solution

From conservation of momentum we know that

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (0.032kg)(230m/s)+(0.15kg)(0m/s)=(0.15kg+0.032kg)v\\7.36+0=0.182v\\v=7.36/0.182\\v=40.44m/s$

Now from the conservation of mechanical energy

$P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\ (9.8m/s^{2} )h=(1/2)(40.44m/s)^{2}\\ (9.8m/s^{2} )h=(817.7m^{2} /s^{2} )\\h=(817.7m^{2} /s^{2} )/9.8m/s^{2}\\ heigth=83.44m$

5 0

3 years ago