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erica [24]
3 years ago
15

When we look up at the stars, we are seeing the future", is this astronomer correct ? Why or why not ?

Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

if the stars connect to a thing, then it describes.

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A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so
hodyreva [135]

Answer:A) Risk(R)= $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R= Risk,

P=Probability of failure

C= Consequence of failure

Mathematically, R=P ×C

10 out of 1000 carbon-steal products failed

Probability of failure= 10/1000 =0.01

The consequence of failure by corrosion given in monetary term =$100,000

Risk of failure = 0.01 × $100,000

R=$1000

4 0
3 years ago
D. What is the net force on the bowling ball rolling lane
3241004551 [841]

Answer:

Friction

Explanation:

3 0
3 years ago
Please help no trolls no links pls
Novosadov [1.4K]
It would move slower/ faster by changing the mass, which would effect the weight and speed.
3 0
3 years ago
Read 2 more answers
Sorry relized it wasnt a good picxD 4 please WILL MARK BRANLIEST
Andrej [43]
I think Y is the answer to number 4
7 0
4 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
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