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3 years ago

Pls help, i really dw fail :(

Physics

2 answers:

mixas84 [53]3 years ago

4 0

Answer:

2.A

3.D

Explanation:

HOPE THIS HELPS!! :)

In-s [12.5K]3 years ago

3 0

Answer:

Using the Newton equation

V²=U² +2as

Given; V=9.2m/s U = 0m/s( because it starts from rest)

S=26.2m.

9.2²=0²+2(26.2)a

a =9.2²/2(26.2)

a=84.6/52.4

a=1.60m/s²

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Which of the following describes density only in terms of metric units?

Alborosie

A) Kilograms per cubic meter. Every other option either contains pounds or feet, which are both units of measurement from the standard system, not the metric system.

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3 years ago

17 points

const2013 [10]

The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>

MV = mv................. Equation 1

<h3>Where:</h3>

M = mass of the cannon
m = mass of the cannonball
V = velocity of the cannon
v = velocity of the cannonball

Make v the subject of the equation.

v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>

M = 500 kg
V = 3 m/s
m = 10 kg.

Substitute these values into equation 2.

v = (500×3)/10
v = 150 m/s.

Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

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2 years ago

A 1.0 kg rock moving at 8.4 m/s will have ______ of kinetic energy

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3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )