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2 years ago

If the net force acting on an object increases by 50 percent, then the acceleration of the object will

1 answer:

8 0

If the net force acting on an object increases by 50 percent, then
the acceleration of the object will also increase by 50 percent.

This answer is not offered among the list of choices.

So the correct response is "D. none of the above"

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

mars1129 [50]

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

8 0

2 years ago

In a scientific test conducted in Arizona, a special cannon called HARP (High Altitude Research Project) shot a projectile strai

Alexus [3.1K]

Answer:

It took the projectile 120 s to reach the maximum height.

Explanation:

Given;

maximum height of the projectile, s = 180 km = 180,000 m

initial speed of the projectile, u = 3 km/s = 3000 m/s

final velocity at maximum height, v = 0

Apply the following kinematic equation for average velocity of the projectile;

$s = (\frac{v+u}{2} )t\\\\(v+u)t = 2s\\\\t = \frac{2s}{v+u} \\\\t = \frac{2*180,000}{0+3,000}\\\\ t = 120 \ s$

Therefore, it took the projectile 120 s to reach the maximum height.

5 0

2 years ago

To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what

neonofarm [45]

T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east

5 0

2 years ago

Read 2 more answers

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil

marishachu [46]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

k' = 160,000 / 4 = 40,000 N/m

b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

To get the frequency, we need to know the angular speed of the car with the following expression:

wo = √k^n / M

M: mass of the wheel and axle, which is 50 kg

k = 40,000 N/m

Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

5 0

2 years ago

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 $kg \cdot cm^2$

Explanation:

Given:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

To Find:

The moment of inertia of Alex's rolling hoop = ?

Solution:

The moment of inertia = $mr^2$

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia = $(0.350)(0.75)^2$

=> moment of inertia = $(0.350)(0.5625)$

=> moment of inertia = $(0.197)$

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 $kg m^2 s^{-1}$

If she pulls her arms in to 0.12 m, her new linear speed is $18.33 m/s^2$

Explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

$L = r \cdot m \cdot v_1$

Substituting the values,

$L = 0.8 \cdot 2.0 \cdot 1.1$

L= 1.76 $kg m^2 s^{-1}$

b. If she pulls her arms in to 0.12 m, what is her new linear speed

$0.12 \cdot 0.8 \cdot v_2 = 1.76$

$0.096 cdot v_2 = 1.76$

$v_2 = \frac{1.76}{0.096}$

$v_2 = 18.33 m/s^2$

6 0

2 years ago