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3 years ago

Iron has more mass than gold. Does this mean that the density of iron is greater

Physics

2 answers:

Strike441 [17]3 years ago

6 0

Answer:

There masses are identical "I hope this helps"

kicyunya [14]3 years ago

3 0

I think it does mean that yeah

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A nation undergoing the transition from undeveloped to industrialized is commonly dependent on this to supply its energy needs.

algol [13]

A) obviously :)))))))))

7 0

3 years ago

Read 2 more answers

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago

The rainbow is a result of a. Diffraction of the light.

suter [353]

Rainbows are caused by the dispersion of light, which itself consists of a combination of refraction and reflection of light around little droplets of water.

Choice C

5 0

3 years ago

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni

nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0

3 years ago

Read 2 more answers