A force of 50 N stretches a string by 4 cm,calculate the elastic constant.

A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre

sesenic [268]

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula : $P(r=3)=^nC_r p^r q ^{n-r}$

$P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166$

Standard deviation = $\sqrt{n \times p \times q}$

Standard deviation = $\sqrt{7 \times 0.15 \times 0.85}$

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

4 0

3 years ago

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Anestetic [448]

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

8 0

3 years ago

Read 2 more answers

A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case

vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction, $\rm \mu_k$ =0.520

Coefficient of static friction, $\rm \mu_s=0.520$

The friction force, F=?

Resolve the force in the inclined plane;

$\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times cos 25 ^0 \\\\ F= 46.2 \ N$

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

5 0

2 years ago

A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l

lianna [129]

For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.

3 0

2 years ago

An airplane propeller is rotating at 1900 rpm (rev/min).

alexandr1967 [171]

Answer:

See explanation

Explanation:

We have to convert to angular velocity in rads-1 as follows;

Angular velocity in rad/s = 2π/60 × 1900 rpm = 199 rad/s

Given that

angular velocity =angle turned /time taken

Time taken = angle turned/angular velocity

Converting 35° to radians we have;

35 × π/180 = 0.61 radians

Time taken = 0.61 radians/199 rad/s

Time taken = 0.0031 seconds

3 0

3 years ago