1. You are given a bottle that contains 4.59 cm of a metallic solid. The total mass of the bottle
is 35.66 g. The empty bottle weighs 14.23 g. What is the density of the solid?​

Explanation:

because a tertiary prevention is when the person is already infected with the disease

Answer:

2400A

Explanation:

R=pL/A

R-Resistance

p-Resistivity=10.0 x 10^-8

L-1.5m

A-Cross sectional Area=0.00045

R=10.0 x 10^-8 x 1.5 ÷ 0.00045

and=0.000333333

I-current

V-potential difference

I=V/R

0.800÷0.000333333

Ans=2400A

Answer: 9.42

Explanation: yes

Answer:

(B) 80 ohms

Explanation:

Since the circuit is a series connection, simply add the two resistances together to get 80 ohms.

Answer:

[tex]0.00195\ \text{m}[/tex]

[tex]0.00293\ \text{m}[/tex]

Explanation:

m = Order = 1

D = Distance between screen and slit = 3.09 m

d = Slit distance = 1.01 mm

[tex]\lambda[/tex] = Wavelength = 639 nm

Distance from the first bright fringe from the central bright fringe is given by

[tex]y=\dfrac{m\lambda D}{d}\\\Rightarrow y=\dfrac{1\times 639\times 10^{-9}\times 3.09}{1.01\times 10^{-3}}\\\Rightarrow y=0.00195\ \text{m}[/tex]

Distance from the first bright fringe from the central bright fringe is [tex]0.00195\ \text{m}[/tex]

Distance from the second dark fringe from the central bright fringe is given by

[tex]y=(m+\dfrac{1}{2})\dfrac{\lambda D}{d}\\\Rightarrow y=(1+\dfrac{1}{2})\dfrac{639\times 10^{-9}\times 3.09}{1.01\times 10^{-3}}\\\Rightarrow y=0.00293\ \text{m}[/tex]

Distance from the second dark fringe from the central bright fringe is [tex]0.00293\ \text{m}[/tex].

Answer:

226.8 kN/m

Explanation:

The work done by the spring, W equals the kinetic energy of the satellite, K

W = K

work done by the spring, W = 1/2kx² where k = force constant and x = extension of spring

kinetic energy of the satellite, K = 1/2mv² where m = mass of satellite = 1060 kg and v = speed of satellite = 3.35 m/s

1/2kx² = 1/2mv²

k = mv²/x²

Also, the spring force F = kx where k = force constant and x = extension of spring.

k = F/x

equation both expressions for k, we have

mv²/x² = F/x

x = mv²/F since F = ma where m = mass of satellite and a = maximum acceleration of satellite = 5.00g and g = 9.8 m/s²

x = mv²/ma = mv²/5.00mg = v²/5.00g

Substituting the values of the variables into the equation, we have

x = v²/5.00g

= (3.35 m/s)²/(5.00 × 9.8 m/s²)

= 11.2225 m²/s²/49 m/s²

=0.229 m

Now k = F/x = 5.00mg/x

Substituting the values of the variables into the equation, we have

k = 5.00mg/x

k = 5.00 × 1060 kg × 9.8 m/s²/0.229 m

k = 51940 kgm/s²/0.229 m

k = 51940 N/0.229 m

k = 226812.23 N/m

k = 226.81223 kN/m

k ≅ 226.8 kN/m

Answer:

C. Solids,Liquids, Gases

Using waves of frequency and pich

Explanation:

There's always been water around us

Answer:

so you have a question

Explanation:

either way,you have a nice day

Answer:

We can write   S = v * t    where S is the distance traveled in time t

S = .12 rad/sec * 30 sec = 3.6 rad     distance traveled

Also 2 * pi * N = S   since there are 2 * pi rad / revolution

So 2 * pi * N = 3.6

N = 3.6 / (2 * pi) = .573 revolutions

Answer:

a.5.85 s b i. 15.7 m/s ii. 21.1m/s c. 61.62 m

Explanation:

(a) when b will overtake a,

Using s = ut + 1/2at² we express the distance moved by each automobile.

Since both automobiles start at rest, their initial speed, u = 0 m/s, a = acceleration of automobile and t = time of travel or overtaking

So s = 0 × t + 1/2at²

s = 0 + 1/2at²

s = 1/2at²

Let time, t be the time the automobile b start, and its acceleration a = 3.6 m/s²

So, s = 1/2 × 3.6 m/s² × t² = 1.8t²

Since automobile a starts 2 seconds earlier, its time of travel is (t + 2) s. Since its acceleration, a = 2 m/s²,

s = 1/2 × 2 m/s² × (t + 2)²

s = (t + 2)²

Since both distances are equal at overtaking

(t + 2)² = 1.8t²

t² + 4t + 4 = 1.8t²

1.8t²- t² - 4t - 4 = 0

0.8t² - 4t - 4 = 0

dividing through by 0.8, we have

0.8t²/0.8 - 4t/0.8 - 4/0.8 = 0

t² - 5t - 5 = 0

Using the quadratic formula to find t,

[tex]t = \frac{-(-5) +/-\sqrt{(-5)^{2} - 4 X 1 X (-5)} )}{2 X 1} \\t = \frac{5 +/-\sqrt{25 + 20} )}{2} \\t = \frac{5 +/-\sqrt{45} )}{2} \\t = \frac{5 +\sqrt{45} )}{2} or \frac{5 -\sqrt{45} )}{2}\\t = \frac{5 + 6.708 )}{2} or \frac{5 - 6.708)}{2}\\t = \frac{11.708 )}{2} or \frac{-1.708}{2}\\t = 5.854 sor -0.854 s[/tex]

t ≅ 5.85 or -0.85

We take the positive answer since t cannot be negative.

So, t = 5.85 s

So, b will overtake a 5.85 s later.

(b) the speed of each automobile at that time,

Using v = u + at for each automobile, where u =initial speed of automobile = 0 m/s (since they both start from rest), a = acceleration of automobile and t = time of travel of automobile

i. Speed of automobile a

For automobile a, its time of travel is (t + 2) s = 5.85 s + 2 s = 7.85 s and its acceleration is 2 m/s²

So, v = u + at

= 0 m/s + 2 m/s² × 7.85 s

= 0 m/s + 15.7 m/s

= 15.7 m/s

ii. Speed of automobile b

For automobile a, its time of travel is t  s = 5.85 s and its acceleration is 3.6 m/s²

So, v = u + at

= 0 m/s + 3.6 m/s² × 5.85 s

= 0 m/s + 15.7 m/s

= 21.06 m/s

≅ 21.1 m/s

c) the location where b overtakes a relative to the traffic signal. -truck

The location where b overtakes a relative to the traffic signal is the distance each automobile moves before overtaking. In section (a), we foundthat distance s = 1/2at² = (t + 2)² = 1.8t²

Using s = (t + 2)²

= (5.85 + 2)²

= 7.85²

= 61.6225 m

≅ 61.62 m

Answer:

please give me brainlist and follow

Explanation:

Gravitational force keeps it attracted to the Earth, and centripetal force keeps it moving in a circle. ... Explain how Newton's third law and his law of universal gravitation are connected. His third law states that every force has an equal opposite force attracted to it, and that force is caused by gravity.

Answer:

[tex]v=11.135m/s[/tex]

Explanation:

From the question we are told that

Mass [tex]m=0.40kg[/tex]

Frictional friction [tex]f=1.2[/tex]

Spring constant [tex]k=800N/m[/tex]

Compressed [tex]x=0.50[/tex]

Generally the energy balance equation is mathematically given as

[tex]1/2kx^2=f*30+m*g*sin \theta*30+1/2m*v^2[/tex]

if [tex](sin=10/30-1/3)[/tex]

[tex]1/2(800)(0.5)^2=(1.2)*30+(0.4)*9.8*1/3*30+1/2(0.4)*v^2[/tex]

[tex]100=36+39.2+0.20v^2[/tex]

[tex]v^2=\frac{24.8}{0.20}[/tex]

[tex]v=\sqrt{124}[/tex]

[tex]v=11.135m/s[/tex]

Answer:

v = C / n     light slows entering a medium of higher index of refraction

ni sin theta i = nr sin theta r    Snell's Law where i refers to incidence and r refers to refraction

sin theta r = (ni / nr) sin theta i

So if the index of refraction is greater than the angle of incidence then the light ray would be refracted towards the normal

Answer:

A. It slows down, and the angle decreases.

Explanation:

got it right, trust

Answer:

33.6 / 96   is the fraction of the temperature between ice and freezing

33.6 / 96 = .35

So C = .35 * 100 = 35 deg C  where the ice point would be 0 deg C

You can convert C to F

F = 9 / 5 * C + 32 = 9/5 * 35 + 32 = 95 deg F

Answer: A cloud of stellar dust from part of a Nebula collapsed causing hydrogen atoms to fuse together

Explanation:

Answer:

r = 0.0173 m = 1.73 cm

Explanation:

Here, the centripetal force of the block will be providing the required breaking tension in the string:

[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]

where,

r = radius = ?

m = mass of block = 0.13 kg

v = tangential spee of block = 4 m/s

T = Breaking Strength = 30 N

Therefore,

[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]

r = 0.0173 m = 1.73 cm

When finding the radius of the string at the point it breaks, the tangential

velocity is assumed to be constant.

Reasons:

The mass of the small block, m = 0.130 kg

Initial radius of the circle of rotation = 0.800 m

Tangential velocity, v = 4.00 m/s

The radius of the path of rotation is reduced as the string is pulled

Breaking strength of the string = 30.0 N

Required:

The radius of the circle when the string brakes

Solution:

[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]

Where;

r = The radius of the circle of rotation

When the string brakes, w have;

Centripetal force = Breaking strength of the string = 30.0 N

Which gives;

[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]

The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]

Learn more here:

https://brainly.com/question/20905151

Answer:

Explanation:

My name is Jeff

Answer:

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

Answer:

it will most likely be the free energy

There actually three assertions here: one in A and two more in B.

All three assertions, as well as the reasoning described in this mess, are all False.

The correct choice from the list is e), but that doesn't even describe the complete Falseness of this disaster.

Answer:

The point will travel a distance of 15708 centimeters in 30 seconds of rotation.

Explanation:

In this case, we see a disk rotating at constant rate, the travelled distance of a point on the outside rim ([tex]s[/tex]), in centimeters, is determined by using this expression:

[tex]s = \omega \cdot r\cdot t[/tex] (1)

Where:

[tex]\omega[/tex] - Angular speed, in radians per second.

[tex]r[/tex] - Radius of the disk, in centimeters.

[tex]t[/tex] - Time, in seconds.

If we know that [tex]\omega \approx 10.472\,\frac{rad}{s}[/tex], [tex]r = 50\,cm[/tex] and [tex]t = 30\,s[/tex], then the travelled distance of the point is:

[tex]s = \omega \cdot r\cdot t[/tex]

[tex]s = 15708\,cm[/tex]

The point will travel a distance of 15708 centimeters in 30 seconds of rotation.