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tester [92]
3 years ago
15

People began to study classical thought wing the Renaissance to​

Physics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

Explanation:

to better themselves and to be good educated people

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig
Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0
3 years ago
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Nata [24]
When the child is moving fastest
6 0
3 years ago
Read 2 more answers
A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 35.0 min
Lubov Fominskaja [6]

Explanation:

Solution:

Let the time be

t1=35min = 0.58min

t2=10min=0.166min

t3=45min= 0.75min

t4=35min= 0.58min

let the velocities be

v1=100km/h

v2=55km/h

v3=35km/h

a. Determine the average speed for the trip. km/h

first we have to solve for the distance

S=s1+s2+s3

S= v1t1+v2t2+v3t3

S= 100*0.58+55*0.166+35*0.75

S=58+9.13+26.25

S=93.38km

V=S/t1+t2+t3+t4

V=93.38/0.58+0.166+0.75+0.58

V=93.38/2.076

V=44.98km/h

b. the distance is 93.38km

6 0
4 years ago
What is the force of friction acting on the crate? I am offering 30 points.
olganol [36]
I think the answer is C


4 0
3 years ago
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