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3 years ago

True or false? When an object deforms, the change of shape is always permanent. false true

1 answer:

8 0

True. Because say if you go to a junk place where you can break anything and you decide to squish a printer with a car and when you go look at the printer it’s all deformed there’s really no way that you can fix it. Lol that was an example

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How many years would take to terraform for Mars

dybincka [34]

Some say one year and a half

8 0

3 years ago

During a tug-of-war, team A pulls on team B by applying a force of 1390 N to the rope between them. How much work does team A do

butalik [34]

Answer:

5115.2 J

Explanation:

Work: This can be defined as product of force and distance. The S.I unit of work is Joules (J).

From the question,

The expression for work is given as

W = F×d................. Equation 1

Where W = Work done by team A, F = Force applied by team A, d = distance moved by team B.

Given: F = 1390 N, d = 3.68 m

Substitute into equation 1

W = 1390×3.68

W = 5115.2 J.

Hence the work done by team A in pulling team B = 5115.2 J

7 0

4 years ago

In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary sec

Pie

Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

8 0

3 years ago

An arrow is launched from a bow with an initial horizontal velocity of 40

Ivan

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

4 0

3 years ago

A ski jumper travels down a slope and leaves

Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

$v_x = 28 m/s$

and the distance covered along the horizontal direction in a time t is

$d_x = v_x t$

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity $u_y = 0$ and constant acceleration $g=9.8 m/s^2$ (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

$d_y = \frac{1}{2}gt^2$

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

$\theta = 36^{\circ}$ below the horizontal

This happens when:

$tan \theta = \frac{d_y}{d_x}$

Substituting and solving for t, we find:

$tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s$

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

$d_x = v_x t = (28)(4.15)=116.2 m$

$d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m$

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m$

7 0

3 years ago