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1 year ago

True or false: An object’s weight can change, but its mass remains the same. True False

Physics

1 answer:

sineoko [7]1 year ago

4 0

The answer is True the mass is the matter of the object while the weight is the gravitational force on the object

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Use your data (attachment) from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving a

vazorg [7]

The cart experiences a frictional force which is directly proportional to its weight. This means that there must be a force applied on the car to balance the forces on the car to produce a net force of 0.
This is in accordance to Newton's first law which states that an object at rest will remain at rest and an object in motion will remain in motion unless an external force acts on it. The force must be a resultant force.
Therefore, the force needed increases with the total weight of the cart as well as with the added mass in a linear manner.

4 0

2 years ago

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

What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be

nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0

3 years ago

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient $\alpha=23.0\times10^{-6}/C^{\circ}$

We need to calculate the volume expansion coefficient

Using formula of volume expansion coefficient

$\beta=3\alpha$

Put the value into the formula

$\beta=3\times23.0\times10^{-6}$

$\beta=0.000069=69\times10^{-6}/C^{\circ}$

We need to calculate the volume

$V= a^3$

$V=4^3$

$V=64\ cm^3$

The change temperature of the cube is

$\Delta T=T_{f}-T_{i}$

Put the value into the formula

$\Delta T=67-19 = 48^{\circ}C$

We need to calculate the increases volume

Using formula of increases volume

$\Delta V=V\beta\Delta T$

Put the value into the formula

$\Delta V=64\times69\times10^{-6}\times48$

$\Delta V=0.212\ cm^3$

Hence, The volume of an aluminum cube is 0.212 cm³.

5 0

2 years ago

Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener

maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0

2 years ago