All categories

4 years ago

What would be a reasonable measurement of normal force if friction force measured 80 Newtons?

1 answer:

4 0

These answers are very good. I made a 95% by using this as my answer key, but unfortunately, one of these is wrong. I don't know which one, all I know is that I got 19/20. Just wanted to throw that out there. :-) Good job, Sadaqasalaam3. And thank you.

You might be interested in

According to the cooling curve, what is the approximate freezing point of the substance?

Oksana_A [137]

Answer:

I do not know which substance you are referring to, but the freezing point of water is 32°F, or 0°C.

3 0

3 years ago

Calculate the percentage of an iceberg submerged beneath the surface of the ocean given that the density of ice is 916.3kg/m3 an

pogonyaev

Answer:

The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%

Explanation:

Let V be the total volume of the iceberg

Let x be the volume of iceberg submerged

According to Archimedes principle,

weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)

Weight of the iceberg = mg= ρ(iceberg) × V × g

Weight of water displaced = ρ(fluid) × x × g

We then have

ρ(iceberg) × V × g = ρ(fluid) × x × g

(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%

Hope this Helps!!!!

6 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

4 years ago

The device or instrument that transforms light energy to electrical energy

WINSTONCH [101]

Answer:

Solar panels

sun light into electricity

3 0

3 years ago