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IRISSAK [1]
2 years ago
12

a body of mass 2kg is released from rest on a smooth plane at an angle of 60 degree to the horizontal.calculate the acceleration

of the body down the plane (g=10ms-2​
Physics
1 answer:
Lelu [443]2 years ago
5 0

Answer:

8.66m/s2

Explanation:

from newton second laws F=ma

the force down the plane is the component force along the plane which is mgsin₩ so there fore

a= gsin₩ = 10sin60= 8.66m/s2

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suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain
IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

Learn more about kinetic energy and potential energy:

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5 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0
2 years ago
(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached
Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

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2 years ago
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Its part of the electromagnetic spectrum. it works by emitting from the sun.
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