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2 years ago

2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction

Physics

1 answer:

Anettt [7]2 years ago

6 0

<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of the brick as 3 kg
The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

We need to importantly note that;
For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.

That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

F = 49 N

Therefore, she must use a force of 49 N

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Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

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The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

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2 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

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2 years ago

What is the acceleration of a car initially traveling at -5m/s and<br> reaching -22m/s in 3s.

lozanna [386]

Answer:

$a=-5.67\ m/s^2$

Explanation:

Given that,

Initial velocity, u = -5 m/s

Final velocity, v = -22 m/s

Time, t = 3s

We need to find the acceleration of the car. The formula of it is given by :

Acceleration,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(-22)-(-5)}{3}\\\\a=-5.67\ m/s^2$

So, the acceleration of the car is $-5.67\ m/s^2$ .

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2 years ago

What is the following correct way to write 2,330,000 In a scientific notation

NISA [10]

Answer:

2.33 × 10^6

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NemiM [27]

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