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Nezavi [6.7K]
3 years ago
8

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

e maximum value of a allowed by these regulations.
Physics
1 answer:
hram777 [196]3 years ago
4 0

maximum allowed value of the speed in roller coaster is given as

v = 20 m/s

now from kinematics we can say

v^2 - v_i^2 = 2 a s

here initial speed will be

v_i = 0

acceleration is due to gravity

a = 9.8 m/s^2

now we can use this to find the height

20^2 - 0^2 = 2 * 9.8* h

400 = 19.6 *h

h = 20.4 m

so maximum allowed height will be 20.4 m

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A block of wood has density 0.500 g/cm3 and mass 2 000 g. It floats in a container of oil (the oil's density is 0.750 g/cm3). Wh
Gennadij [26K]

Answer:

2,666.67cm^3

Explanation:

All we need to do in this problem is divide the mass of the wooden block to the oil's density.

2000/0.750 ≈ 2,666.67cm^3

Best of Luck!

6 0
2 years ago
For the ride to be comfortable, the magnitude of acceleration must not exceed 32 m/s2. What is the fastest constant speed that a
VMariaS [17]

Answer:

the maximum possible constant speed is  8 m/sec  

Explanation:

from the image, Given that

r(t) = (2t, t²,t²/3), -5 ≤ t ≤ 5  

Given that the curvature K(t)  = 2 / ( t² + 2)²  

note that t² + 2 ≥ 2  

(t² + 2)² ≥ 4

1 / (t² + 2)² ≤ 1/4

2 / (t² + 2)² ≤ 1/2

Also note that k(0) = 1/2

The normal component of acceleration satisfies aN = kv²

where v = ║v(t)║is the speed of the roller coaster.

The maximum possible normal component of acceleration is 32

so, aN ≤ 32 every where on the track

aN = kv² ≤  1/2v² ≤ 32

v² ≤ 64

Therefore, the maximum possible constant speed is  8 m/sec  

7 0
2 years ago
A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.050
BartSMP [9]

Answer:

Explanation:

Given that,

Slender rod

Length of rod=80cm=0.8m

Mass of slender rod=0.12kg

Sphere Bob at one end

Mass M1=0.02kg

Sphere Bod at the other end

Mass M2 =0.05kg

Linear speed of mass 2 at the lowest point

We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.

So, at the lowest point,

∆U = M2•g•y2 + M1•g•y1

Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is -40cm(-04m) from the midpoint

∆U = M2•g•y2 + M1•g•y1

∆U=0.05•9.81•(-0.4) + 0.02•9.82•0.4

∆U=-0.1962+0.07848

∆U=-0.11772 Nm

Now, the moment of inertia of the rod is given as

I=∫r²dm

dm=2pdr

I= 2p∫r²dr

I= 2 × 0.12/0.8 ∫r²dr; from r=0 to 0.4

I=0.3 [r³/3] from r=0 to 0.4

I= 0.3 [ 0.4³/3 -0] ,from r=0 to 0.4

I=0.3 × 0.02133

I=0.0064kg/m².

calculating of inertia of the end masses.

I(1+2)=Σmr² = (m1+m2)r²

I(1+2)=(0.02+0.05)0.4²

I(1+2)=0.07×0.4²

I(1+2)=0.0112 kg/m²

Now, the Energy of the masses due to angular velocity is given as

K.E=½ (I + I(1+2))w²

K.E=½(0.0064+0.0112)w²

K.E= 0.0088w²

Using conservation of energy

The potential energy is equal to the kinetic energy of the system

K.E=P.E

0.0088w²=0.11772

Then, w²=0.11772/0.0088

w²=13.377

w=√13.377

w=3.66rad/s

Then, the relationship between linear velocity and angular velocity is given by

v=wr

v=3.66×0.4

v=1.463m/s

The required linear speed is 1.46m/s approximately

8 0
3 years ago
HELP ASAP!!! PLEASE!!
prohojiy [21]

Answer:

Explanation:

The weaker force is pulling to direction East(450 newton)

the net force is (500-450=50 newton) and will move toward west ( in 500 newton direction)

6 0
3 years ago
How do objects become negatively charged using the contact method
Kamila [148]

If a negative object is used to charge a neutral object, then both objects become charged negatively. In order for the neutral sphere to become negative, it must gain electrons from the negatively charged rod. A metal sphere is electrically neutral. It is touched by a positively charged metal rod.

7 0
3 years ago
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