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3 years ago

Why is it difficult to cut using a pair if of scissors with blunt blades

1 answer:

8 0

Its not sharp enough to cut through whatever your trying to cut. Imagine trying to cut paper with your index finger and your middle finger. Doesn't work does it?

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A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/

Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

$T = ma + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2$

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

5 0

2 years ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

katrin2010 [14]

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

5 0

3 years ago

For rectilinear motion the relationship between displacement, s, velocity, v, acceleration, a, and time, t, s = s0 + v0t + (1/2)

stiv31 [10]

The rectilinear motion is given as:

s = s0 + v0 t + (1/2) a t^2

We can see that the variables stand for:

s = final distance

s0 = initial distance

v0 = initial velocity

t = time

a = acceleration

<span>We can see that all these variable are always constant EXCEPT acceleration. The rectilinear motion can only be applied if and only if acceleration is CONSTANT.</span>

6 0

3 years ago

At t= 10 s, a particle is moving from left to right with a speed of 5.0 m/s. At t=20 s, the particle is moving right to left wit

Shkiper50 [21]

Answer

given,

at time t = 10 s moving from left to right speed of particle = 5 m/s

at t = 20 s moving from right to left speed of particle is = 8 m/s

a) acceleration

$a = \dfrac{\Delta V}{\Delta t}$

$a = \dfrac{-8-(5)}{20-10}$

a = -1.3 m/s²

b) using equation of motion

v = u + at

5 = u - 1.3 × 10

u = 5 + 13

u = 18 m/s

c) v = u + at

0 = 18 - 1.3t

$t = \dfrac{18}{1.30}$

t = 13.85 s

4 0

3 years ago

A cannon tilted up at a 30 angle fires a cannon ball at 80 m/s from atop a 10-m-high fortress wall. what is the ball's impact s

lora16 [44]

To find the velocity of the bullet just before reaching the ground, we use the formula.
Vy = Vosen (30 °) + gt (1).
For that we need to know the time it takes for the bullet to reach the ground from the moment it is fired.
For that we use the formula:
Xy = Xo + Vosen (30 °) * t + 0.5gt ^ 2 (2)
where:

Vy = Component of the speed in the y direction (vertical)
Vo = initial velocity of the bullet at the exit of the cannon
g = acceleration of gravity
Xy = vertical component of the position.
Xo = initial position.
t = time.

To find the time it takes for the bullet to reach the ground, we make Xy = 0 and clear t.
then it would be:
0 = 10 + 80sen (30 °) * t +0.5 * (- 9.8) t ^ 2.
The solution to that equation is:

t = 8.4 seconds.

We substitute that time in equation (1) and clear Vy.
We have left:

Vy = 80 * sin (30 °) + 9.8 * (8.4)
Vy = 42.38 m / s

7 0

2 years ago