All categories

3 years ago

HELP WILL GIVE BRAINLIEST IF CORRECT

Physics

2 answers:

ki77a [65]3 years ago

5 0

Answer:

A. a a a a a a a a a aa a a a a a a a. aa a a

PolarNik [594]3 years ago

3 0

Answer:

the answer is 1.35. Have a nice day!!

You might be interested in

What is the work done in lifting 60 kg of blocks to a height of 20m

erastovalidia [21]

Answer:

The answer is 1200

Explanation:

5 0

2 years ago

Convert 9/4 hours into minutes

Eduardwww [97]

Am guessing;

9/4 × 60/1 =135

5 0

3 years ago

How are theories and laws connected

Anni [7]

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

4 0

3 years ago

Read 2 more answers

A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, αα-4. The numbers indicate the

alexdok [17]

Answer: E = 5.80*10^-13 J

Explanation:

Given

We use the law of conservation of momentum to solve this

Momentum before breakup = momentum after breakup

0 = m1v1 + m2v2

0 = 238m * -2.2*10^5 + 4m * v2

0 = -523.6m m/s + 4m * v2

v2 * 4m = 523.6m m/s

v2 = 523.6 m m/s / 4m

v2 = 130.9*10^5 m/s

v2 = 1.31*10^7 m/s

Using this speed in the energy equation, we have

E = 1/2m1v1² + 1/2m2v2²

E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²

E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]

E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)

E = 9.56*10^-15 + 5.7*10^-13

E = 5.80*10^-13 J

3 0

4 years ago

A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an

elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)= $(2.4-x^2)\hat{i}N$

Initial position of the block=x=0

Initial velocity of block= $v_i=0$

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy= $K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0$

Work energy theorem:

$K_f-K_i=W$

Where $K_f=$ Final kinetic energy

$K_i$ =Initial kinetic energy

$W=Total work done$

Substitute the values then we get

$K_f-0=\int_{0}^{2}F(x)dx$

Because work done= $Force\times displacement$

$K_f=\int_{0}^{2}(2.4-x^2)dx$

$K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}$

$K_f=2.4(2)-\frac{8}{3}=2.13 J$

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy = $K=2.4x-\frac{x^3}{3}$

When the kinetic energy is maximum then $\frac{dK}{dx}=0$

$\frac{d(2.4x-\frac{x^3}{3})}{dx}=0$

$2.4-x^2=0$

$x^2=2.4$

$x=\pm\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2x$

Substitute x= $\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2\sqrt{2.4}$

Substitute x= $-\sqrt{2.4}$

$\frac{d^2K}{dx^2}=2\sqrt{2.4}>0$

Hence, the kinetic energy is maximum at x= $\sqrt{2.4}$

Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by

$K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx$

$k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}$

$K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J$

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0

4 years ago