Which is a common disease involving the sores on the esophagus, stomach, and/or small intestine?

If a jet is travelling horizontally at 100m/s at a height of 500m above the ground and a jet drops a bomb to the ground. Where d

Westkost [7]

The bomb strike the ground relative to the point at 1km . B

<h3>How to determine the distance</h3>

Using the equation

h = 1/2 gt^2

500 = 1/2 * 10* t^2

500 = 5t^2

t = √500/5

t = √100

t = 10seconds

To find the distance,

Distance = velocity * time

Distance = 100 ÷ 10

Distance = 1000m = 1km

Therefore, the bomb strike the ground relative to the point at 1km . B

Learn more about projectile distance here:

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8 0

2 years ago

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0

Leya [2.2K]

Explanation :

When a ball is tossed from an upper storey window off a building, the height of the object as a function of time is given by :

$h(t)=ut+\dfrac{1}{2}at^2$

Here, u = 8 m/s

The ball strikes at an angle of 20 degrees below the horizontal. We need to find the time taken by the ball to reach a point 10 meters below the level of launching such that, h (t) = 10 m and a = g = 9.8

$10=ucos(20)t+\dfrac{1}{2}\times 9.8\times (t)^2$

$10=0.93\times 8t+4.9t^2$

$10=7.51t+4.9t^2$

After solving the above equation, t = 0.855 seconds

So, the ball will take 0.855 seconds to reach a point 10 m below the level of launching. Hence, this is the required solution.

3 0

3 years ago

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At wha

dezoksy [38]

a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always $9.8 m/s^2$ downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

$a=g=-9.8 m/s^2$

downward (acceleration due to gravity). So, we can use the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u = 40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s$

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

$v=u+at$

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

$t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s$

c)

To solve this part, we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

$0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0$

which gives two solutions:

t = 0 (initial instant)

$u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s$

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

$v=u+at$

where

u = +40.0 m/s is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

$t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s$

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

$F=mg$

where m is the mass of the boulder and $g$ the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

$F=ma$

Combining the two equations, we get

$ma=mg\\a=g$

So, the acceleration of the boulder is $g=9.8 m/s^2$ downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is $-9.8 m/s^2$ , so this graph is a straight horizontal line.