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A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon

notsponge [240]

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

= 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

5 0

4 years ago

Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged

MArishka [77]

Answer:

A) E = 0N/C

B) 0i + 0^^j

C) F = 0N

D) 0^i + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

$E=k\frac{rQ}{(r+R^2)^{3/2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

$E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}$

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

$F=qE=q(0N/C)=0N$

D) The direction of the force is 0^i + 0^j

6 0

3 years ago

1) A particle move along the x axis, as x=tsquare-2t+1(SI unit). (a) The functions of its velocity and accelerator vector agains

Bond [772]

Answer:

$\text{I believe this is calculus, no? But I believe that there should be a y-component of position}\\\text{But based off of the information provided, here goes:}\\\text{(a)}\\x'(t)=v(t)=\frac{d}{dt}t^2-2t+1\\v(t)=2t-2\text{ is the velocity vector}\\v'(t)=a(t)=\frac{d}{dt}2t-2\\a(t)=2 \text{ is the acceleration vector}\\\text{(b)}\\\text{Average Velocity }=\frac{x_{f}-x{i}}{t_{f}-t_{i}}\\=\frac{1-1}{2}=0, \text{based off of the information provided, the average velocity in the first 2 seconds}: 0m/s$

8 0

3 years ago

If I was driving at 50 miles an hour and hit a car at a stop light, how would there car react after it is hit?

MA_775_DIABLO [31]

Answer:

The answer is C.move forward rapidly until stopped.

Explanation:

50 miles an hour is a very high speed. The momentum transfer to the car at stop light will move the car forward rapidly until its brake is able to stop it.

4 0

3 years ago

Read 2 more answers

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

nlexa [21]

Answer:

Explanation:

hi

5 0

3 years ago