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A merry go round exerts a force of 1000 N on a rider on the

SVEN [57.7K]

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

$F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r$

Where;

m = The mass of the person

v = The velocity of the person

$F_c$ = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for the centripetal force, we have;

$F_c$ = m × ω² × r

1000 = 76.45 × (2·π/15)² × r

r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

3 0

3 years ago

A piston motion moves a 25-kg hammerhead vertically down 1 m from rest to a velocity of 50 m/s in a stamping machine. What is th

Alja [10]

Answer:

31.005 KJ

Explanation:

We are given that

Mass of hammerhead=25 kg

Initial velocity,u=0

Final velocity,v=50 m/s

h=-1 m

h'=0

We have to find the change in total energy of the hammerhead.

Change in total energy= $E_2-E_1=m(u_2-u_1)+\frac{1}{2}m(v^2-u^2)+mg(h-h')$

Using the formula

Change in internal energy of hammerhead= $m(u_2-u_1)=0$

Change in total energy= $\frac{1}{2}(25)(50)^2+25\times 9.8(-1)$

Where $g=9.8m/s^2$

Change in total energy= $31005 J=\frac{31005}{1000}=31.005 KJ$

1 KJ=1000 J

3 0

3 years ago

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw

pshichka [43]

Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

What is the magnitude of its velocity just before it strikes the ground ?

The parameters given are:

Height H = 100m

Since the ball is thrown from a top of a building, initial velocity U = 0

Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

Substitute all the parameters into the formula

V^2 = 2 × 9.8 × 100

V^2 = 200 × 9.8

V^2 = 1960

V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

6 0

3 years ago

A 6.0-kg ball is sliding to the right at 25.0m/s and strikes a second ball (15-kg) that is initially at rest head-on. After the

katrin [286]

Answer:

15 m/s to the right

Explanation:

Let's say right is positive and left is negative.

Momentum is conserved:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(6.0 kg) (25.0 m/s) + (15 kg) (0 m/s) = (6.0 kg) (-12.5 m/s) + (15 kg) v₂

v₂ = 15 m/s

5 0

3 years ago

If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.

motikmotik

Answer:

d

Explanation:

8 0

3 years ago