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3 years ago

What is the average velocity of Runner C for the first 15 seconds? Question 5 options: 1.3 m/s -2.7 m/s -1.3 m/s 2.7 m/s

Physics

2 answers:

kogti [31]3 years ago

7 0

Answer:

2.7 m/s

Explanation:

Took the test and got 100%

kipiarov [429]3 years ago

3 0

Answer:

Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two.

v¯¯¯=(v+u)2

Where:

v¯¯¯ = average velocity

v = final velocity

u = initial velocity

The average velocity calculator solves for the average velocity using the same method as finding the average of any two numbers. The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.

Explanation:

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In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

3 years ago

1) A record is spinning at the rate of 25 rpm. If a ladybug is sitting 10 cm from the

Sladkaya [172]

<h2>distance = 523 cm</h2>

Explanation:

( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s

= 5/12 rev/sec

( b ) The definition of frequency is the number of rotations per second .

Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz

( c ) The tangential speed is v = angular velocity x radius of rotation

The angular velocity ω = 2π x n , where n is the number of rotations per second

Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec

The linear velocity = angular velocity x distance from center of record

Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec

Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad

Linear displacement = angular displacement x distance from center of record

= 50π/3 x 10 = 500π/3 = 523 cm

8 0

2 years ago

How many significant figures does the following number have? <br><br> 9876

Agata [3.3K]

Answer:

Four

Explanation:

7 0

2 years ago

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:

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