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2 years ago

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Calculate the KE of a 3.1 kg toy cart that moves at 4.8 m/s. Calculate the KE of the same cart at twice the speed.

1 answer:

lisov135 [29]2 years ago

4 0

When v= 4.8 m/s:

E.K=mv²/2
=3,1(4.8)²/2
=35.712J

when v= 2(4.8):

E.K=mv²/2
=3.1(9.6)²/2
=142.848J

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A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?

vovikov84 [41]

We can solve the problem by using Newton's second law of motion:
$F=ma$
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
$a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2$

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>

5 0

2 years ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

8 0

2 years ago

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s

AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight $W = 150$ N

Length of rope $r = 4.1$ m

Initial speed of ball $v = 5.5 \frac{m}{s}$

For finding the tension in the rope,

First find the mass of rod,

$mg = 150$ ( $g = 9.8 \frac{m}{s^{2} }$ )

$m = \frac{150}{9.8}$

$m = 15.3$ kg

Tension in the rope is,

$T = mg + \frac{mv^{2} }{r}$

$T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}$

$T = 262.88$ N

Therefore, the tension in the rope is 262.88 N

7 0

2 years ago

How do magnetic trains work and what advantages do they have compared to regular trains

ad-work [718]

Answer:

magnetic trains works at the principle of repel on of the advantage is that they are fast and dont really need diesel

7 0

2 years ago

Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori

LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0

3 years ago