When you see yourself in a plane mirror, the image is always:

Stells [14]

Solar energy, because of solar panels, hydroelectricity from a dam, or fossil fuels. fossil fuels would be the most ideal because it is more commonly used

3 0

3 years ago

What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?

kipiarov [429]

Answer:

$\boxed {\boxed {\sf a= -7 \ m/s^2}}$

Explanation:

Acceleration is the change in velocity over time.

$a= \frac {v_f-v_i}{t}$

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

$v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s$

Substitute the values into the formula.

$a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}$

Solve the numerator.

$a= \frac { -35 \ m/s}{5 \ s}$

Divide.

$a= -7 \ m/s/s$

$a= -7 \ m/s^2$

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0

2 years ago

A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight

Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is 7.61 x 10⁻⁶

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

m is mass supported by the steel
g is acceleration due to gravity
A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

ΔL is change in length
L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0

1 year ago

A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy

MrRa [10]

The kinetic energy is $\frac 1 2 m v^2$ and the height of the building doesn't matter at all.

$E = \frac 1 2 m v^2 = \frac 1 2 (2)(5)^2 = 25$ joules

8 0

2 years ago

A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th

vivado [14]

Answer:

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

$\dfrac{20}{30}\times 30=20\ cm$

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

7 0

2 years ago