All categories

3 years ago

Is the speed of the object is zero when it's at rest ?

Physics

1 answer:

NARA [144]3 years ago

4 0

Yes! Although the object has potential energy, the object is not moving and therefore has no speed. :)

You might be interested in

A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if

Zanzabum

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

$m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm$

$v=-2.5\times 21=-52.5\ cm$

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

3 0

2 years ago

Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1

anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

ΔL/Lo = αΔT

solve for ΔL

ΔL = αLoΔT

Substitution

ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

ΔL = 0054 m = 5.4 cm

7 0

3 years ago

If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.

Marizza181 [45]

V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance.

7 0

2 years ago

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

morpeh [17]

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

$PE=mgh$

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

$PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

6 0

2 years ago

Read 2 more answers

Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?

IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity = 3.8km/h

Unknown:

time = ?

Solution:

Velocity is displacement divided by time.

Velocity = $\frac{displacement}{time}$

Displacement = velocity x time

Input the parameters:

8 = 3.8 x time

Time = $\frac{8}{3.8}$ = 2.1s

The time taken is 2.1s

6 0

2 years ago