Someone just help me

Over the past 150 years, what has happened to the amount of forest cover in Minnesota?

Airida [17]

I think D. None of the above

I think that it

4 0

3 years ago

PLZ HELPPPPP!! i'll give brainliest

kati45 [8]

Answer:

12N to the right.

Explanation:

There is a force of 12N upwards and a force of 12N downwards: these cancel out.

Assign a negative value to forces towards the left, and a positive value to the forces towards the right: -3N and +15N

Combine them: -3N+15N = 12N

The net force has a magnitude of 12N, and since our answer was positive, it acts towards the right.

5 0

3 years ago

Which object will sink in freshwater, which has a density of 1.0 g/cm3?

Pani-rosa [81]

Answer:

Object 2, which has a density of 1.9 g/cm3, since it has more density than freshwater.

6 0

3 years ago

Read 2 more answers

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago

How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?

lukranit [14]

<span>work =V*Q =12*50*10^-6

The total work done will be equal to

work = V.Q

which means

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>

8 0

3 years ago