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liq [111]
3 years ago
11

amy and ellie left school at the same time amy lives farther away then ellie but she and ellie arrived at theure homes at the sa

me time compare the girls speed
Physics
1 answer:
velikii [3]3 years ago
8 0

Army and Ellie left school at the same time.

Army lives farther than Ellie.

They arrived at their homes at the same time.

This means that Army travels faster than Ellie.

Time \frac{distance}{speed}

Since the time is constant here, increasing the distance directly increases the speed.

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A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t
damaskus [11]
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
3 0
3 years ago
The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the g
Reptile [31]
<span>H(t) = -16t^2 + vt + s
</span><span>Part A:
</span>Using the given data:

H(t)= -16*t² + 60*t + 82;

Part B:
Put H(t)=0
0<span>= -16*t² + 60*t + 82;</span>
Use the quadratic formula to find t.
See the attachment...'t' is replaced with 'x'.

3 0
3 years ago
In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a
pickupchik [31]

Answer:

time of collision is

t = 0.395 s

h = 5.63 m

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

v_1 = 0

similarly initial speed of the object which is projected by spring is given as

v_2 = 16.2 m/s

now relative velocity of object with respect to ball

v_r = 16.2 m/s

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

d = v_r t

6.4 = 16.2 t

t = 0.395 m

Now the height attained by the object in the same time is given as

h = v_2 t - \frac{1}{2}gt^2

h = 16.2(0.395) - \frac{1}{2}(9.81).395^2

h = 5.63 m

so they will collide at height of 5.63 m from ground

4 0
3 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0
3 years ago
4application of energy from water
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5 0
3 years ago
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