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3 years ago

12

A 100 kg roller coaster car is located at the top of a hill where the height is 20 m. What is the velocity when it reaches the b

ottom of the hill?

2 answers:

galina1969 [7]3 years ago

4 0

The velocity of the car at the bottom of the hill=19.8 m/s

mass= 100 kg

height= h=20 m

Velocity at the bottom=V

Using the law of conservation of energy

potential energy at top = kinetic energy at the bottom

m g h= 1/2 m v²

g h= 1/2 v²

9.8 (20)=1/2 V²

V=19.8 m/s

Thus the velocity of car at the bottom of hill=19.8 m/s

djyliett [7]3 years ago

4 0

Answer: the final velocity is 19.8 m/s

Explanation:

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WHAT IS THE NET FORCE REQUIRED TO GIVE AN AUTOMBILE OF MASS 1600KG AN ACCELERATION OF 4.5M/S2?

Fed [463]

Answer:

The required net force has a magnitude of 7200 N

Explanation:

Use Newton's 2nd Law to obtain the answer:

$F_{net}= m\,*\,a\\F_{net}=1600 \,*\,4.5 \,= 7200\,\,N$

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4 years ago

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

¡¡first convert revolutions per second to

katovenus [111]

Answer:

a. 5.23 m/s² b. 44.23 N

Explanation:

a. What is the centripetal acceleration of the hammer?

The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.

Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.

So a = rω²

= 1.4 m × (3.74 rad/s)²

= 5.23 m/s²

b. What is the tension in the chain?

The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.

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6 0

3 years ago

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Inessa [10]

Answer:

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Explanation:

4 0

4 years ago

Read 2 more answers

Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~

Rama09 [41]

Answer:

34.70 N

Explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

= 43 (0.7) + 26 (-0.7) + 27 (-0.3)

= 3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

= 43 (0.6) + 26 (-0.6) + 27 (0.9)

= 34.5

so, F~ = $\sqrt{3.8^2 + 34.5^2}$

= 34.70 N

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3 years ago