All categories

3 years ago

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

Physics

1 answer:

NARA [144]3 years ago

4 0

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:

- A uniform motion with constant velocity along the horizontal direction

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

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

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

$u_y=0$ is the initial vertical velocity (the missile is thrown horizontally)

t is the time of flight

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

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s$

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

$v_x = 45 m/s$

Therefore, the distance covered in a time t is

$d=v_x t$

and by substituting t = 3.91 s, we find the horizontal range of the missile:

$d=(45)(3.91)=176 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

You might be interested in

An adult human is 60 percent water; a third of this water is in extracellular fluid, and 20 percent of extracellular fluid is in

Yanka [14]

Answer:

32 pounds

Explanation:

The amount of water in the 200 pound person is

200 * 60% = 200*0.6 = 120 pounds

Of the 120 pounds, a third of this is extracellular fluid, the amount of extracellular fluid is

120 / 3 = 40 pounds

20 % of this is in the blood, which amounts to

40 * 20% = 40 * 0.2 = 8 pounds

The rest is interstitial fluid, which is

40 - 8 = 32 pounds

5 0

3 years ago

A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted thro

aalyn [17]

Answer:

The percentage power lost in the transmission line if the voltage not stepped up is 50%.

Explanation:

Given that,

Current = 60 A

Voltage = 120 V

Resistance = 1.0 ohm

We need to calculate the power

Using formula of power

$P=I\times V$

Where,I =current

V = voltage

Put the value into the formula

$P=60\times120$

$P=7200\ W$

We need to calculate the percentage power lost in the transmission line

If the voltage is not stepped up

Then, the power loss

$P'=I^2\times R$

Put the value into the formula

$P'=(60)^2\times1$

$P'=3600\ W$

The percentage power loss P''

$P''=\dfrac{P'}{P}\times100=\dfrac{3600}{7200}\times100$

$P''=50\%$

Hence, The percentage power lost in the transmission line if the voltage not stepped up is 50%.

5 0

2 years ago

Help with physics projectile motion

BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0

3 years ago

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

2 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago