What electrical appliance was invented in 1910

Svet_ta [14]

B long board because of the length

7 0

3 years ago

A spring has a spring constant of 81 N · m−1. What is the force (in N) required to do the following? (Enter the magnitude.) (a)

DochEvi [55]

Explanation:

It is given that,

Spring constant, k = 81 N/m

We need to find the force required to :

(a) Compress the spring by 6 cm i.e. x₁ = 6 cm = -0.06 m

It can be calculated using Hooke's law as :

F = - k(-x₁)

$F=81\ N/m\times 0.06\ m$

F = 4.86 N

(b) Expand the spring by 17 cm i.e. x₂ = 17 cm = +0.17 m

So, F = -kx₂

$F=-81\ N/m\times 0.17\ m$

F = -13.77 N

Hence, this is the required solution.

8 0

3 years ago

A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq

Snowcat [4.5K]

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block $W=\frac{1}{2}kx^2$

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block $W=-\frac{1}{2}kx^2$

Thus option d is correct

6 0

3 years ago

Suppose your vehicle is moving 12m/s as it crosses the 7m line. It took 6 seconds to get to the 7m line. What was its accelerati

Inga [223]

Answer:

3.61 m/s²

Explanation:

Given:

v = 12 m/s

x = 7 m

x₀ = 0 m

t = 6 s

Find:

a

x = x₀ + vt - ½ at²

7 m = 0 m + (12 m/s) (6 s) - ½ a (6 s)²

a = 3.61 m/s²

8 0

3 years ago

Suppose electric power is supplied from two independent sources which work with probabilities 0.4, 0.5, respectively. if both so

sattari [20]

Answer:

The answers to the questions are as follows

a) k = 0, P = 0.3

k = 1, P = 0.5

k = 0, P = 0.2

b) The probability that enough power will be available is 0.5.

Explanation:

To solve the question we write the parameters as follows

Probability that the first power source works = P(A) = 0.4

Probability that the second power source works = P(B) = 0.5

When both sources are supplying power we have the probability = 1

If non of them is producing the probability = 0

a) The probability that exactly k sources work for k=0,1,2 is given by

For k = 0, probability = (1- P(A))× (1- P(B)) = 0.6 × 0.5 =0.3

Therefore the probabilities that exactly 0 source work = 0.3

for k = 1 we have the probability = P(A)(1-P(B)) + P(B)(1-P(A)

= 0.4(1-0.5)+0.5(1-0.4) = 0.2 + 0.3 = 0.5

The probabilities that exactly 1 source work = 0.5

for k = 2 we have the probability given by = P(A) × P(B) = 0.4 × 0.5 = 0.2

Therefore the probability that exactly 2 sources work = 0.2

b) The probability that enough power will be available is

0 × P(k = 0) + 0.6 × P(k = 1) + 1 × P(k = 2)

0 × 0.2 + 0.6 × 0.5 + 1 × 0.2 = 0.5

The probability that enough power will be available is 0.5.

3 0

3 years ago