Which of Newton's motion laws BEST explains HOW a rock accelerates when it is dropped off a bridge?

A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle

Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $\alpha =2.538 \ rad/s^2$

Explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is $r = 0.69 \ m$

The radius of gyration is $k_G = 0.4\ m$

The angle is $\theta = 47^o$

The force which the massless bar is subjected to $F = 22.5 \ N$

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

$I_a = I_G + M* r^2$

Here $I_G$ is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

$I_G = M * k_G$

=> $I_a = k_G* M + M* r^2$

=> $I_a =0.4 * 6.9 + 6.9 * 0.69^2$

=> $I_a =6.045 \ kg \cdot m^2$

Generally the torque experienced by the wheel is mathematically represented as

$\tau = F * cos (47)$

=> $\tau = 22.5 * cos (47)$

=> $\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}$

Generally this torque is also mathematically represented as

$\tau = I_a * \alpha$

=> $15.34 = 6.045 * \alpha$

=> $\alpha =2.538 \ rad/s^2$

4 0

3 years ago

A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is

andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0

3 years ago

A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2

asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

$\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}$

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

$\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s$

4 0

3 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

2 years ago

Julian and Joshua each maintain a constant speed as they run laps around a 400-meter track. In the time it takes Julian to compl

Marysya12 [62]

Answer:

the time Joshua travels 1 mile is 12.5 min

Explanation:

Let's start by finding the distance traveled on each lap,

Let's reduce everything to the SI system

R = 400 m

d = 1 mile (1609 m / 1 mile) = 1609 m

L = 2 pi R

L = 2 pi 400

L = 2513 m

Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took

lap Julian = 2/3 turn Joshua

Let's calculate what distance is the same for both of them since they are on the same track

1 lap = 2513 m

d. Julian = 2/3 2513 m

d Julian = 1675 m distance Joshua

Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m

t Julian = 1675/1609 12

t = 12.5 s

Since this is the distance Joshua travels, this is the time Joshua travels 1 mile

5 0

3 years ago

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