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Gekata [30.6K]
4 years ago
11

Lexy throws a dart with an initial velocity of 25 m/s at an angle of 60° relative to the ground. what is the approximate vertica

l component of the initial velocity?
Physics
2 answers:
Serjik [45]4 years ago
7 0
An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.
Greeley [361]4 years ago
4 0

Answer:

21.7

Explanation:

Because you need to round it

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A car was traveling at 20m/s and slammed on the brakes causing the car to decelerate at a rate of -2.5 m/s^2 before coming to a
Taya2010 [7]

Answer: 160 m

Explanation: First find time or t using the formula for acceleration:

a = vf - vi / t

So t is equal to:

t = vf - vi / a

= 0 m/s - 20 m/s / -2.5 m/s²

= 8 s

To find the distance use the formula for velocity:

v = d / t

Derive for d:

d = v x t

= 20 m/s x 8 s

= 160 m

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3 years ago
HELP ME, I'M BEING TIMED
m_a_m_a [10]

Answer:

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Explanation:

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3 years ago
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17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.
creativ13 [48]

a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

a)

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s

b)

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

K_i = U_f

\frac{1}{2}mv^2=mgh

where:

K_i is the initial kinetic energy

U_f is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

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

h is the maximum height reached by the projectile

And solving for h, we find

h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m

c)

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

v=u-gt

where:

u = 16.7 m/s is the initial velocity

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

t is the time

The projectile reaches the maximum height when the vertical velocity becomes zero, so when v = 0. Therefore, substituting,

0=u-gt\\t=\frac{u}{g}=\frac{16.7}{9.8}=1.70 s

So, the total time of flight is

T=2t=2(1.70)=3.40 s

d)

The motion of a projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction  

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

First we have to analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

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

where

s = -27 m is the vertical displacement of the apple

u_y=u sin \theta = (16.7)(sin 42^{\circ})=11.2 m/s is the initial vertical velocity

t is the time if flight

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

Substituting we have:

-27=11.2t - 4.9t^2\\4.9t^2-11.2t+27=0

which has two solutions:

t = -1.46 s (negative, we discarde)

t = 3.75 s (this is our solution)

Now we can analyze the horizontal motion: the projectile moves horizontally with a constant velocity of

v_x = u cos \theta = (16.7)(cos 42^{\circ})=12.4 m/s

So, the distance it covers during its fall is given by

d=v_x t=(12.4)(3.75)=46.5 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0
4 years ago
A liquid enters one end of a tube, where the diameter is
Kamila [148]

Answer:

 v_2=3.2\ m/s

Explanation:

given,

diameter of inlet = 11.2 cm

                     r₁ = 5.6 cm

                     r₁ = 0.056 m       ∵ 1 cm = 0.01 m  

speed of inlet = 5 m/s

diameter of outlet = 17.5 cm

                    r₂ = 8.75 cm

                    r₂ = 0.0875 m       ∵ 1 cm = 0.01 m  

speed of outlet = ?

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_2= \dfrac{r_1^2}{r_2^2} v_1

 v_2= \dfrac{0.056^2}{0.0875^2} v_1

 v_2= 0.64 v_1

 v_2= 0.64\times 5

 v_2=3.2\ m/s

Velocity of liquid at the outlet of the tube is equal to 3.2 m/s

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3 years ago
What would a Munsell scale be used for
vekshin1
The answer to this question is A."Characterizing soil color"
-I hope this is the answer you are looking for.
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3 years ago
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