Question 10 of 10

a 12000.0-kg car is traveling at 19 m/s. the driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic

tensa zangetsu [6.8K]

What I don’t get it ?

6 0

2 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

2 years ago

half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds

Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %

3 0

3 years ago

A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?

Marizza181 [45]

According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons

6 0

3 years ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to 22.5N.

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that the force tending to lift the load off the ground (vertical force) is equal to 22.5N.

Have a nice day!

8 0

3 years ago