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Vsevolod [243]
3 years ago
6

a runner runs 2.88 m/s north. she accelerates .350 m/s ^2 at a -52.0 degree angle at the point in the motion where she is runnin

g directly east what is her displacement y
Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

uh

Explanation:

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Mgh=mv²/2

v=√2gh=√2·9.8·1.25=4.95m/s
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3 years ago
How many normal modes of oscillation or natural frequencies does each if the following have: (
Vadim26 [7]
<span>Each of these systems has exactly one degree of freedom and hence only one natural frequency obtained by solving the differential equation describing the respective motions. For the case of the simple pendulum of length L the governing differential equation is d^2x/dt^2 = - gx/L with the natural frequency f = 1/(2π) √(g/L). For the mass-spring system the governing differential equation is m d^2x/dt^2 = - kx (k is the spring constant) with the natural frequency ω = √(k/m). Note that the normal modes are also called resonant modes; the Wikipedia article below solves the problem for a system of two masses and two springs to obtain two normal modes of oscillation.</span>
6 0
3 years ago
Simplify 6.25 − 8.<br><br>please help
tatyana61 [14]

6.25 - 8 = -1.75

Hope this helps

-AaronWiseIsBae

7 0
3 years ago
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal
kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both  spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0
3 years ago
What are the physical characteristics of sound waves
OlgaM077 [116]
•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
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