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3 years ago

A rock is thrown horizontally at 10.0m/s off of a 150m sheer cliff. What is the range?

Physics

2 answers:

inn [45]3 years ago

7 0

Answer:20

Explanation:

sattari [20]3 years ago

3 0

Answer:

55.3 m

Explanation:

First, find how long it takes to land.

Given (in the y direction):

Δy = 150 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(150 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 5.53 s

Given (in the x direction):

v₀ = 10.0 m/s

a = 0 m/s²

t = 5.53 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (10.0 m/s) (5.53 s) + ½ (0 m/s²) (5.53 s)²

Δx = 55.3 m

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Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$ :

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

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3 years ago

5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in

prisoha [69]

The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

c. V2 equals V1.

d. V2 equals (V1)/4.

e. V2 equals 4V1.

Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

ρ1•V1•A1 = (ρ1)/2•V2•A2

Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

5 0

2 years ago

A nearby star has a parallax of 0.2 arc seconds. What is its distance?

ruslelena [56]

Answer:

5 parsecs

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3 years ago

a canon is firing horizontally from the high wall of a castle located in the midst of a flat plain. if i drop a canon ball from

dsp73

D=44.13, Horizontal velocity is unimportant. The time it takes the cannon ball to fall to the ground is the key to determining the height of the cliff. where that cannon is present.

The speed of any projectile travelling along a Horizontal velocity is known as the horizontal velocity. When a particle or object is launched into the air at an angle other than 90 degrees, it moves along the trajectory path and changes the shape of the curve to a parabolic one.

the speed at which velocity changes over time. Due to its magnitude and direction, acceleration is a vector quantity. The first derivative of velocity with respect to time or the second derivative of position with respect to time are further examples. This is called acceleration.

Distance to the ground is d = 1/2gt^2,

where g is the acceleration rate of gravity (9.80665 m/s^2)

and t = 3 secs.

d = .5×9.80665×9 = 44.13 m.

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10 months ago

What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.

soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0

2 years ago