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2 years ago

Evaluate the expression a+b when a=23 and b=45 . Write in simplest form

Physics

1 answer:

Viefleur [7K]2 years ago

6 0

If a = 23 and b = 45

a + b = ?

23 + 45 = 68

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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in

Novay_Z [31]

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

μ(s) is coefficient of static friction
m is mass of the block
g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

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4 0

1 year ago

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 $m/s^2$

Substituting

$520 = u*223\\ \\u = 2.33 m/s$

The constant velocity of car a = 2.33 m/s

b) We have $s = ut+\frac{1}{2} at^2$

s = 520 m, t = 223 seconds, u =0 m/s

Substituting

$520 = 0*223+\frac{1}{2} *a*223^2\\ \\ a = 0.0209 m/s^2$

Now we have v = u+at, where v is the final velocity

Substituting

v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 $m/s^2$

7 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

3 years ago

What is calculated by the equation KE = 1/2mv2?

Anit [1.1K]

Kinetic Energy is calculated by KE= 1/2mv2

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2 years ago

The stored chemical energy in fuels is MOST OFTEN converted to _________ and _________ energy through combustion.

Vlad1618 [11]

I believe is c) light and electrical

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3 years ago