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mina [271]
3 years ago
8

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c

rater. If the volcanic rocks were launched at an angle of 40° with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?
Physics
1 answer:
Nata [24]3 years ago
7 0

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

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The refraction of light makes a swimming pool seem <u>shallower</u>.

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6 0
1 year ago
DETERMINE THE BCD EQUIVALENT OF 1100111110101001
Solnce55 [7]

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6 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

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

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
A student is observing a pendulum swinging back and forth. Each time it swings, the pendulum bob reaches a lower maximum, and ev
vodomira [7]

Answer:

the pendulum loses momentum and stops because of gravity and wind resistance. it does not violate the law of conservation of energy because it is not gaining any more momentum than what it had started with

Explanation:

7 0
2 years ago
Read 2 more answers
A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
3 years ago
Read 2 more answers
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