All categories

3 years ago

Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.

1 answer:

3 0

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

You might be interested in

Plz do all plz i will give brainest and thanks to best answer do it right

lara31 [8.8K]

Answer:

I'm not really sure but I think it's choice a

5 0

3 years ago

Read 2 more answers

Dwayne ‘The Rock’ Johnson needs to escape from the fourth floor of a burning building (in a movie). He ties a rope around his wa

ZanzabumX [31]

Answer:

Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s

Explanation:

Let's start out with finding the force acting downwards because of the mass of 'The Rock':

Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N

Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

787.58 = 63 x acceleration

acceleration = 12.501 m/s

Finally, we can use the motion equation:

$v^2 - u^2 = 2*a*s$

here u = 0 m/s (since initial speed of the desk is 0)

a = 12.501 m/s

and s = 10 m

Solving this we get:

$v^2 - 0 = 2 * 12.501 * 10$

$v = 15.812 m/s$

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.

5 0

3 years ago

A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067

Vesna [10]

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

$P = mv$

$m = \dfrac{P}{v}$

Where, P = momentum

v = speed

Put the value into the formula

$m = \dfrac{2.8}{187}$

$m = 0.015\ kg$

$m = 15\ g$

Hence, The mass of the bullet is 15 g.

4 0

3 years ago

How is lithium fluoride and lithium chloride similar?

Jet001 [13]

Is this science because I am suck at it

3 0

3 years ago

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric

ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ = -ε²/(R + r)²

cancelling out the common variables

2R = R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0

3 years ago