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Juli2301 [7.4K]
1 year ago
13

A car is going 43m/s and travels 14 m, how long did it take?

Physics
1 answer:
Burka [1]1 year ago
6 0

Answer:

Time = 0.3256 sec

Explanation:

Velocity = displacement / time

43=14/t

So the time =14/43

t=0.3256 sec

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Which graph shows a negative acceleration
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Answer:

The velocity-time graph shows a line with a negative (downward) slope (negative acceleration); the line is located in the positive region of the graph (positive velocity).

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In foot x-ray, what is the distance/FFD and Why?
MakcuM [25]

Answer and Explanation:

FFD is the distance between the film on which the image is obtained and the center of the anode tube. The magnification and resolution of the image depends on the FFd By varying the FFD we can change the magnification and resolution of the image. The standard FFD is about 100 centimeters.

New studies have found that by changing the FFD to 130 cm the radiation dosage reduces while the image quality remains practically the same.

5 0
1 year ago
A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.
eimsori [14]

1) x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

The initial data of the projectile are:

y_0 = 12 m is the initial height

v_0 = 32 m/s is the initial speed of the projectile, so its components along the x- and y- directions are

v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

0=12+21.8t-4.9 t^2

Solving the equation with the formula, we have:

t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

v_y(t)= v_{0y} -gt

where

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

Substutiting v_y(t)=0 and solving for t, we find

t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m

3 0
1 year ago
How do you calculate the average speed of an object?
tensa zangetsu [6.8K]

Answer:

The average speed of an object is the total distance traveled by the object divided by the elapsed time to cover that distance. It's a scalar quantity which means it is defined only by magnitude. A related concept, average velocity, is a vector quantity.

Explanation:

4 0
1 year ago
Read 2 more answers
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
1 year ago
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