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3 years ago

A 150 g egg is dropped from 3.0 meters. The egg is

Physics

1 answer:

Lynna [10]3 years ago

5 0

Answer:

9.2 N, with significant figure rounding (2 s.f.)

Explanation:

This problem can be solved using momentum. The following equation relates momentum (mass & velocity) with force and time:

Note that where v is the final velocity and v₀ is the initial velocity. Δv just means change in velocity.

Mass of the egg is 150 g, but we need to convert to kilograms if we want to use Newtons as a unit. 150 g is equal to 0.15 kg. since 1000 g = 1kg.

m = 0.15 kg

The dropped from 3.0 meters is irrelevant as the question tells us the initial velocity of the egg: 4.4 m/s before it hits the ground.

v₀ = 4.4 m/s [down]

When it comes to a stop, the egg will have a velocity of 0.

v = 0 m/s

The time it takes for the egg to stop is 0.072 seconds.

Δt = 0.072 s

Therefore, if down is positive, then

We round to two significant figures since every quantity has two sig. figs.

We only care about the magnitude, not direction. The answer is 9.2 N.

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5.0

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The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop

nalin [4]

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration $a= -4.20 m/s^2$

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

$0 =24.0 +(-4.20)\times t$

$t = \dfrac{-24.0}{-4.20}$

$t=5.71\ sec$

Hence, The time is 5.71 sec.

5 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

____________________ when objects collide, the total initial momentum equals the total final momentum

Alexus [3.1K]

The law of conservation of linear momentum says that when objects collide, the total initial momentum equals the total final momentum.
The momentum of a body is given by the product of mass and velocity of the object.

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4 years ago

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According to the Universal Law of Gravitation, every object attracts every other object in the universe. Why can’t you feel the

inessss [21]

Mars is officially frank gallagers son

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3 years ago

The question is in the attachment!!

Bingel [31]

Explanation:

|F|= B/2

B/2 = √(A²+B²+2ABcosθ) ------(1)

Since the resultant of A and B is perpendicular to Vector A

tan90°= BSinθ/(A+BCosθ)

(A+BCosθ)=0

Cosθ=-A/B ----(2)

Using equation (1)

B/2 = √(A²+B²+2ABcosθ)

B/2 = √(A²+B²+2AB×-A/B)

B/2=√(A²+B²-2A²)

B/2=√(B²-A²)

B²/4=B²-A²

A²=B²-B²/4

A²=3B²/4

A=√3B/2

Using equation (2)

Cosθ=-A/B

Cosθ=-[√3B/2]/B

Cosθ=-√3/2

θ= cos^-1 (-√3/2)

θ= 150°

7 0

3 years ago