What is the position and kind of image produced by the lens below?

The velocity of a car increases by 40 m/s in 80 seconds. What is the acceleration of the car

yan [13]

Given that the velocity of a car increases by 40 m/s in 80 seconds, the acceleration of the car will be given by:
a=(final velocity-initial velocity)/(time)
thus;
final velocity=40 m/s
initial velocity=0
time=80 seconds
hence;
a=(40-0)/80
=0.5m/s^2

7 0

3 years ago

What is the value of x?<br><br> 6x−2(x 4)=12

Ymorist [56]

Answer: The value of x is -6.

Explanation:

To calculate the value of 'x', we need to solve each function happening inthe equation.

The equation provided to us is $6x-2(4x)=12$

To solve this, we will multiply 4x with 2 and then subtract the like terms and finally, we evaluate the value of 'x'.

$6x-8x=12\\\\-2x=12\\\\x=\frac{12}{-2}\\\\x=-6$

Hence, the value of x will be -6.

5 0

3 years ago

Read 2 more answers

At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

3 years ago

The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal

viktelen [127]

Answer:

a. $F=7.83\times 10^{-51} N$

b.Attractive

Explanation:

We are given that

$F=\frac{GM_1M_2}{R^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Mass of an electron, $M_1=9.11\times 10^{-31} kg$

Mass of proton, $M_2=1836\times 9.11\times 10^{-31} kg$

Distance between electron and proton,R= $3.602nm=3.602\times 10^{-9} m$

$1nm=10^{-9} m$

a.Substitute the values then we get

$F=\frac{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1836\times 9.11\times 10^{-31}}{(3.602\times 10^{-9})^2}$

$F=7.83\times 10^{-51} N$

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

4 0

3 years ago

How does the toaster serve as an example of Convection energy transfer?

blsea [12.9K]

Check bing for the answer

3 0

3 years ago