T or F A red-hot object is hotter than one that is while-hot?

A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical

DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ $P.E=1000-\frac{1}{2}mv^2$

$=1000-\frac{1}{2}\times 52\times (6)^2$

$=1000-26\times 36$

$=1000-936$

$=64 \ J$

7 0

3 years ago

If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a

PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s $w_o$ = 12 rad/s

Angular Velocity at time 0.15s $w_f$ = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( $w_f$ - $w_o$ ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0

3 years ago

A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25

Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in

4 -in.-thick

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

= 66/ (П* (d₂²-d₁²)/4

=66/ (3.142* (12²-8²)/4

= 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

= 66* cos 25/ (П* (d₂²-d₁²)/4

=59.82/ (3.142* (12²-8²)/4

= 59.82/62.84 = 0.952kips/in²

shearing stress = tangential stress /2

= T/2 = 0.952/2 = 0.476 kips/in²

8 0

3 years ago

When does a rubber band, which has been shot at a wall, have the most potential energy?

Yuri [45]

D When it is stretched ready to shoot at the wall

4 0

3 years ago

Read 2 more answers

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

3 years ago