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2 years ago

A gannet is a seabird that fishes by diving from a great height

1 answer:

3 0

For constant acceleration along a given direction, we can relate acceleration, velocity and position with the following equation that doesn't involve time:
$v^{2}= v_{0}^{2}+2a(x- x_{0})$
In this equation x is the final position, which we take to be 0. Also the initial velocity Vo is zero. Thus the equation simplifies to
$v^{2}= 2a(- x_{0})$
Putting in v=32m/s, a=-9.81m/s^2 gives
$32^{2}= 2(-9.81)(- x_{0})
$
$\frac{32^{2}}{2*9.81}= x_{0}=52.19m$

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The position (in radians) of a car traveling around a curve is described by Θ (t) = t 3 - 2t 2 - 4t + 10 where t (in seconds). W

FromTheMoon [43]

Answer: $\alpha =30-4=26 rad/s^2$

Explanation:

Given

Position of a car is given by

$\theta =t^3-2t^2-4t+10$

and angular speed is $\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

angular acceleration is

$\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}=\alpha$

$\alpha =6t-4$

at $t=5 s$

$\alpha =30-4=26 rad/s^2$

4 0

2 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

babunello [35]

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

7 0

2 years ago

Calculate the average velocity in m/y of a tectonic plate that has travelled 9000 km to the south in 60 million years

Gekata [30.6K]

The distance covered by the tectonic plate, in meters, is
 $d=9000km=9\cdot 10^6 m$ 
The time taken for the tectonic plate to cover this distance is equal to
 $t=60 mil.y=60 \cdot 10^6 y$ 
Therefore, the average velocity of the tectonic plate is the ratio between the distance covered and the time taken:
 $v= \frac{S}{t} = \frac{9\cdot 10^6m}{60 \cdot 10^6y} =0.15 m/y.$

8 0

2 years ago

A 64.5 kg person steps off a 129 kg rowboat with a force of 34.0 N. What is the force that is applied to the person by the rowbo

Arada [10]

34.0 N which is the force that applied to the person by rowboat.

3 0

2 years ago

The speed of sound in air changes with the temperature. When the temperature T is 32 degrees Fahrenheit, the speed S of sound is

dezoksy [38]

Answer and Explanation:

A. We have temperature t = 32

Speed of sound, s = 1087.5

As t increases by 1⁰f speed increases by 1.2

So that

S = 1088.6

T= 33⁰f

We have 2 equations

1087.5 = k(32) + c

1088.6 = k(33) + c

Subtracting both equations

(33-32)k = 1088.6-1087.5

K = 1.1

b.). S = kT + c

1087.5 = 32(1.1) + c

Such that

C = 1052.3

Therefore

S = 1.1(t) + 1052.3

C.). S = 1.1t + 1052.3

We make t subject of the formula

T = s/1.1 - 1052.3/1.1

T = 0.90(s) - 956.3

D. This means that We have temperature to rise by 0.90 whenever speed is increased

8 0

2 years ago