Any followers for me please i have solved 25 question thats too much please

The weight of an iron block is 8.0±0.3? and is placed on a wooden base of area, 3.5±0.2? 2. Calculate the pressure exerted by th

vesna_86 [32]

Answer:

Explanation:

Units...Nothing is accurate without UNITS. 8.0±0.3 lbs,? Newtons? 3.5±0.2 m²? ft²? cm²? km²? furlong²? fathoms²?

your answer will be 8.0±0.3 / 3.5±0.2 units / units²

8 0

2 years ago

Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry

Phantasy [73]

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

3 0

3 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 0

3 years ago

A piston is filled with gas. When the pressure is increasing, what is happening to the volume? Assume that all other properties

Kryger [21]

The correct answer is letter C. Volume is decreasing. For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume<span> are </span>inversely proportional<span>. </span>

3 0

3 years ago

A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the

pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²

W(1 to 2) = 11.25 J

W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0 to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

6 0

2 years ago

Any followers for me please i have solved 25 question thats too much please​

Any followers for me please i have solved 25 question thats too much please