What is the net force necessary for a 1.6 x 103 kg car to accelerate forward at 2.0<br> m/s?

Distance is 80 seconds is 8 what is the average speed?

True [87]

Answer:

speed =distance /time

speed =80/8

speed =10m/s

8 0

2 years ago

Energy is released during the fission of pu-239 atoms as a result of the

pav-90 [236]

<h2>Answer: process of converting matter into energy</h2><h2></h2>

Nuclear fission consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the bombardment with neutrons to make it unstable. In this process that takes place in the atomic nucleus, neutrons, gamma rays and <u>large amounts of energy are emitted. </u>

Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.

This means fission is a process in which energy is released by the separation of the components of the nucleous of the atom.

In other words:

<h2>Matter is converted to energy .</h2>

6 0

2 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

4 0

3 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = +8.4\mu C$

$q_2 = +5.6 \mu C$

force between two charges is given as

$F = 0.66 N$

now we have

$F = \frac{kq_1q_2}{r^2}$

$0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}$

$r = 0.8 m$

so it is separated by 80 cm distance

3 0

2 years ago

Read 2 more answers

The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving

nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

$F_f = \mu_s (mg)$

where

$\mu_s = 0.5645$ is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

$F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N$

So, the applied force must be greater than this value.

3 0

3 years ago

What is the net force necessary for a 1.6 x 103 kg car to accelerate forward at 2.0 m/s?