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2 years ago

What is the difference between an object’s speed and its acceleration? NO LINKS

1 answer:

5 0

Answer: Hello! An objects speed is constant and has the units meters per second (m/s); thus, it does not change overtime. Acceleration is a rate of change where the speed does either increase or decrease overtime from its inital value; its units are meters per second second (m/s/s). I hope that helps!

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A cheerleader lifts his 79.4 kg partner straight up off the ground a distance of 0.945 m before releasing her. the acceleration

Oksi-84 [34.3K]

To find out how much work he has done, we must first calculate force using the force formula (F= Mass*Acceleration). In this case, mass is 79.4 and acceleration is the gravitational constant of 9.8m/s, plugging this into the formula we find that force is 778.12Newtons. Next, we need to multiply force by the distance to get the amount of energy used to lift his partner once. Which is 778.12 * .945 = 735.32. Finally, we need to multiply 735.32 by the number of times he lifts his partner, 33, to get 735.32 * 33 to find that the energy he has expended 24,265.56 Joules of energy.

5 0

2 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

3 years ago

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the

dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

$S=\frac{1}{2}gt^2$

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s$

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

$v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s$

8 0

3 years ago

7. What does the slope of a position-time graph represent?

allochka39001 [22]

Answer:

a. Velocity

Explanation:

The slope of the tangent line on a position-time graph is the instantaneous velocity.

7 0

3 years ago

A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by

Margarita [4]

Thank you for posting your question here at brainly. Feel free to ask more questions.
The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds . 

Hope my answer would be a great help for you.

3 0

3 years ago