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3 years ago

A rock is thrown downwards from the edge of the Grand Canyon. With

Physics

1 answer:

Olin [163]3 years ago

3 0

Answer:

Vf = 28 m/s

Explanation:

In order to find the final velocity of the rock, we will use the 3rd equation of motion. The third equation of motion for vertical direction is written as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height dropped = 40 m

Vf = final velocity of the rock = ?

Vi = Initial Velocity of the rock = 0 m/s (since, rock was initially at rest)

Therefore,

(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²

Vf = √(784 m²/s²)

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You push downward on a trunk at an angle 25° below the horizontal with a force of If the trunk is on a flat surface and the coef

Sophie [7]

Complete question is;

You push downward on a trunk at an angle 25° below the horizontal with a force of 750N. if the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Answer:

The most massive trunk is about 81.3 kg

Explanation:

I've attached a free body diagram that depicts this question.

Where;

N = normal force on the trunk

m = mass of the trunk

W = weight of the trunk = mg

F = static frictional force

Using equilibrium of force in vertical direction, we obtain;

N = W + 750 Sin25

N = mg + 750 Sin25 - - - - (eq 1)

Now, we are given that Coefficient of static friction: μ = 0.61

static frictional force is given by the formula;

F = μN

Since N = mg + 750 Sin25, we now have;

F = (0.61) (mg + 750 Sin25) - - - (eq 2)

Along the horizontal direction, for the trunk to move, force equation must be;

F = 750 Cos25

Thus, we now have;

750 Cos25 = 0.61(mg + 750 Sin25)

g = 9.81.

So,we now have ;

750 Cos25 = 0.61(m(9.81) + 750Sin25)

750 × 0.9063 = 0.61(9.81m + (750 × 0.4226))

Divide both sides by 0.61;

(750 × 0.9063)/0.61 = 9.81m + 316.95

1114.3 = 9.81m + 316.95

1114.3 - 316.95 = 9.81m

797.35 = 9.81m

m = 797.35/9.81

m = 81.3 kg

7 0

3 years ago

A small marble attached to a massless thread is hung from a horizontal support. When the marble is pulled back a small distance

s2008m [1.1K]

Answer:

f' = 2 f

Explanation:

The frequency of the pendulum that swings in simple harmonic motion is given by :

$f={2\pi}\sqrt{\dfrac{l}{g}}$

Where

l is the length of pendulum

g is the acceleration due to gravity

If the length of the thread is increased by a factor of 4, such that, l' = 4 l, let f' is the new frequency such that,

$f'={2\pi}\sqrt{\dfrac{l' }{g}}$

$f'={2\pi}\sqrt{\dfrac{4l}{g}}$

$f'=2\times {2\pi}\sqrt{\dfrac{l}{g}}$

f' = 2 f

So, the new frequency of the pendulum will become 2 time of initial frequency. Hence, the correct option is (b) "2f"

3 0

3 years ago

A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel befor

Alja [10]

Answer:

Δx = 39.1 m

Explanation:

Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)$

where vf is the final velocity (0 in our case), v₀ is the initial velocity

(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

traveled since the brakes are applied.

Solving (1) for Δx, we have:

$\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)$

7 0

3 years ago

The ends of a magnet where thee forces are the strongest are called magnets

ladessa [460]

They are called magnets poles.

7 0

3 years ago

Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

$I_{disk} = \frac{1}{2} mR^2$

HOOP:

$I_{hoop} = mR^2$

SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

4 0

3 years ago