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nataly862011
11 days ago
12

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

water land 2.5m away?
Physics
1 answer:
inna [2.2K]11 days ago
8 0
The velocity of water can be decomposed into its vertical and horizontal components:
v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta

The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:
y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t
To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5
-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3
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When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup
Sav [2226]

Answer:

W_f = 148.17J

Explanation:

The friction created between the tire and the ground generates thermal energy as force is applied during skidding.

The mentioned force relates to half the impact on the rear tire, resulting in a calculated normal force of,

N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N

The work executed is determined by the frictional force and the distance covered,

W_f = fd = \mu_k Nd

Where \mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)

The thermal energy produced from the work done is,

W_f = 148.17J

3 0
19 days ago
A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to
inna [2205]

Answer:

The canyon measures 50.176 meters deep.

Explanation:

The student drops a rock from the rim of the canyon, requiring us to ascertain the depth of the canyon—name how far the ground is below the cliff.

The data we have:

Time = t = 3.2 s

Initial velocity = v_{i} = 0 m/s

Gravitational acceleration = g = 9.8 m/s²

Height = h =?

According to the second equation of motion

h = v_{i}t + \frac{1}{2}gt^{2}

Given the initial velocity is zero, the right-hand side of the equation simplifies to zero

h = \frac{1}{2}gt^{2}

h = (0.5)(9.8)(3.2)²

h = 50.176 m

This calculation indicates that the rock dropped a distance of 50.176 meters to reach the canyon's base. Thus, the canyon depth is 50.176 meters.
5 0
10 days ago
A 10 cm wide box is held between two springs in a 1 m gap on a frictionless surface. The left spring has a natural length of 80
Maru [2355]

Answer:

The distance measures x =0.291 \ m

Explanation:

According to the problem statement,

The box's width is b = 10 \ cm = \frac{10}{100} = 0.10 \ m

There is a gap of length l = 1\ m

The first spring's natural length is y = 80 \ cm = \frac{80 }{100} = 0.8 \ m

The spring constant for the first spring is k_1 = 200 N/m

The second spring has a natural length of z = 90 \ cm = \frac{90}{100} = 0.9 \ m

The second spring's spring constant is k_2 = 350 \ N/m

We denote the distance from the center of the box to the left edge as x.

At equilibrium,

The force exerted by the first spring is

F_1 = k_1 * (0.8 -x)

while the force from the second spring is

F_2 = k_2 * [ 0.9 - (0.9 -x)]

Thus, at equilibrium,

F_1 = F_2

Substituting values gives us

k_1 * (0.8 -x) = k_2 * [ 0.9 - (0.9 -x)]

which leads to

200 * (0.8 -x) = 350 * [ 0.9 - (0.9 -x)]

resulting in

160 -200x) = 350x

and finally,

160 =550x

this simplifies to

x =0.291 \ m

6 0
15 days ago
You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th
Yuliya22 [2420]

Answer:

The y-component of the position vector for the car is 670m/s.

The x-component of the acceleration vector is -3 and the y-component equals 40.

Explanation:

The car's displacement vector corresponding to the velocity

\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s

is derived from the integration of the velocity.

By integrating \boldsymbol{v}, we obtain the displacement vector \boldsymbol{d}:

\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j}  )

Assuming the vehicle's starting position is

\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})

the displacement at time t would then be

\boldsymbol{d(t)}= \boldsymbol{r+d}

\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j}  )}

At the moment t=10s, we find

\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j}  )}m

The y-component of the position vector for the car is 670m/s.

The acceleration vector can be calculated as the derivative of the velocity vector:

\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})

At t=10s, it becomes

\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2

The x-component of the acceleration vector is -3 and the y-component is 40.

5 0
1 month ago
If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don't have time t
Sav [2226]

Complete Question

Standing up quickly from a lying position may lead to dizziness or faintness. This occurs because the blood vessels cannot dilate sufficiently to adjust for the sudden drop in blood pressure. If your brain is 0.4 m above your heart when you are upright, what is the difference in blood pressure between these two points? The density of blood plasma is approximately 1025 kg/m3, and the typical systolic blood pressure at the heart is 120 mm of Hg (which is equal to 0.16 atm, 16 kP, or 1.6 × 104 N/m2).

Answer:

The pressure at the brain is P_b = 89.872 \ mm \ of \ Hg

Explanation:

It is generally represented mathematically as

                  P = \rho gh

By substituting 1025 kg/m^3 for \rho (the density), 9.8 m/s^2 for g (acceleration due to gravity), and 0.4m for h (height)

We can express the pressure difference between the brain and the heart as

              P = 1025 * 9.8 *0.4

                  = 4018 N/m^2

The blood pressure at the heart is given as

               P_h=120 mm of Hg = 120 * 133 = 1.59*10^3Pa

Now, the calculated pressure at the brain becomes

                 P_b = P_h - P

                     = 1.596*10^4 - 4018

                     = 11982 N/m^2

                      P_b= \frac{11982}{133} = 89.872 \ mm \ of \ Hg

   

     

3 0
8 days ago
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