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2 years ago

What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air

Physics

1 answer:

madam [21]2 years ago

8 0

Hi there!

We can use the kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

$v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}$

Substitute in the given values:

$v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}$

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Answer:

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3 years ago

Tina drove 3 miles north and 4 miles west to get to her grandmother’s house. Find the distance and displacement.

TEA [102]

Answer:

distance = 3 + 4 = 7 mi

displacement is √(3² + 4²) = 5 mi

Explanation:

6 0

2 years ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from

kodGreya [7K]

Answer:

$ax = 2.60m/s^{2}$ , $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

Where $(vx)f^{2}$ is the final velocity, $(vx)i^{2}$ is the initial velocity, $ax$ is the acceleration and $\Lambda x$ is the distance traveled.

Equation (1) can be rewritten in terms of ax:

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$ (2)

Since the plane starts from rest, its initial velocity will be zero ( $(vx) = 0$ ):

Replacing the values given in equation 2, it is gotten:

$ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}$

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

So, The acceleration of the plane is $2.60m/s^{2}$

Now that the acceleration is known, the next equation can be used to find out the time:

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

5 0

3 years ago

The gravitational acceleration on Earth is 9. 8 m/s2. What is the weight of a car on Earth (to the nearest whole number) if it h

astraxan [27]

Hi there!

Recall the equation for weight.

$W = Mg$

W = Weight (N)

M = Mass (kg)
g = acceleration due to gravity (m/s²)

The weight of an object depends upon its MASS and the strength of the GRAVITATIONAL field. We can solve for weight:

$W = 1360 * 9.8 = \boxed{13,328 N}$

6 0

2 years ago