Một vật giao động điều hoà với chu kì T = 3,14s và biên độ A=1m tại thời điểm vật qua VTCB có giá trị là

What are the SI units for acceleration?<br> a. m/s<br> b. m/s/s<br> c. N<br> d. kg

ozzi

M/s^2 is the correct answer

7 0

3 years ago

A runner accelerates toward the finish line.

Brut [27]

C the runners feet pushing against the ground describes the acceleration toward the finish line

6 0

3 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0

4 years ago

Read 2 more answers

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s$

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where $\Delta p$ is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

$\Delta t=10.0 ms =0.010 s$ is the duration of the collision

Substituting,

$F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N$

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0

4 years ago

if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?

Katena32 [7]

A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

Learn more about harmonics here brainly.com/question/17315536

#SPJ10

7 0

2 years ago