Which of the following jobs does not require good swimming ability

Why is accelration negative when velocity is positive.

Sindrei [870]

Answer:

An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

Explanation:

8 0

3 years ago

if heat energy is transferred from direct contact between a warm and a cold object it has been transferred by what ?

Katarina [22]

When heat energy is transferred from direct contact between a warm and a cold object , it is known as heat transfer by conduction.

In conduction, the heat transfer takes place within an object or between two objects in contact until the temperature becomes uniform. this kind of heat transfer continues until the temperature at two ends between which the heat transfer take place , becomes equal. Heat transfer takes place from point at high temperature to point at lower temperature.

5 0

3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

A motor cycle travelling at 10 m/s accelerates at 4m/s(squared) for 8s.

Sever21 [200]

10 x 4^2 = 160 / 8..

V = 20m/s...

...x 8 = 100 miles,meters, metric what ever m stands for after 8 seconds.

This is my guess since the problem says 4m/s^2

V= distance/ ST (traveled/used)

8 0

3 years ago

Unit 5 activity 2 I need answer

saul85 [17]

Answer:

Can you explain more in detail what it is?

Explanation:

6 0

3 years ago