What is the connection between change in momentum and impulse

HELP PLS! :/

zlopas [31]

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

7 0

3 years ago

What instrument is normally used to measure atmospheric pressure?

Anuta_ua [19.1K]

B. Barometer is used to measure atmospheric pressure.

6 0

3 years ago

In a solar system far, far away the sun's intensity is 400 W/m2 for an inner planet located a distance R away. What is the sun's

kobusy [5.1K]

Answer:

I₂ = 25 W / m²

Explanation:

Intensity is defined as the relationship between power and area

I = P / A

the power emitted by the sun is constant

P = I A

for the two points of interest

I₁ A₁ = I₂ A₂

energy is distributed on the surface of a sphere

A = 4π R²

I₁ R₁² = I₂ R₂²

I₂ = $\frac{R_1^2}{R_2^2} \ I_1$

let's calculate

I₂ = $\frac{R^2}{(4R)^2} \ 400$

I₂ = 25 W / m²

7 0

2 years ago

A 25.0-kg box is released on a 27° incline and accelerates down the incline at 0.30 m/s2. Find the friction force impeding its m

mr Goodwill [35]

Answer:

(A) Frictional force will be 103.7276 N

(B) Coefficient of friction $\mu _k=0.4751$

Explanation:

We have given

mass of the box m = 25 kg

Angle $\Theta =27^{\circ}[/text]Acceleration [tex]a=0.3m/sec^2$

Apply Newton's second law of motion ,

Vertically :

$F_N$ = mg cosθ

= $25\times 9.8\times COS27^{\circ}=218.296N$

Horizontally :

mg sinθ - F = ma

A)

frictional force F= mg sinθ - ma

= $25\times 9.8sin27^{\circ}-25\times 0.3=103.7276N$

B)

Coefficient of friction $\mu _k=\frac{F}{F_N}=\frac{103.7276}{218.296}=0.4751$

8 0

3 years ago

I

Virty [35]

Answer:

The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.

Explanation:

From Classic Electrostatic Theory, we find that electric field ( $E$ ), measured in newtons per coulomb, is defined by the following equation:

$E = \frac{F}{q_{O}}$ (1)

Where:

$F$ - Electrostatic force, measured in newtons.

$q_{O}$ - Electric charge, measured in coulombs.

In addition and supposing that phenomena occurs between particles, electrostatic force is modelled after the Coulomb's Law:

$F = \frac{k\cdot q\cdot q_{O}}{r^{2}}$ (2)

Where:

$k$ - Electromagnetic constant, measured in newton-square meters per square coulomb.

$q$ , $q_{O}$ - Electric charges, measured in coulomb.

$r$ - Distance between particles, measured in meters.

By applying (2) in (1), we get the following definition of electric field:

$E = \frac{k\cdot q}{r^{2}}$ (3)

Then, we observe that electric field is inversely proportional to the square of the distance. The following relationship is therefore constructed:

$\frac{E_{2}}{E_{1}} = \left(\frac{r_{1}}{r_{2}} \right)^{2}$ (4)

If we know that $E_{1} = 10\,\frac{N}{C}$ , $E_{2} = 2.50\,\frac{N}{C}$ and $r_{1} = 1\,m$ , then the distance is:

$\left(\frac{E_{2}}{E_{1}} \right)\cdot r_{2}^{2}= r_{1}^{2}$

$r_{2}^{2}=\left(\frac{E_{1}}{E_{2}} \right)\cdot r_{1}^{2}$

$r_{2} = r_{1}\cdot \sqrt{\frac{E_{1}}{E_{2}} }$

$r_{2} = (1\,m)\cdot \sqrt{\frac{10\,\frac{N}{C} }{2.50\,\frac{N}{C} } }$

$r_{2} = 2\,m$

The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.

4 0

3 years ago