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Drupady [299]
3 years ago
12

Two rollerbladers face each other and stand at rest on a flat parking lot. Tracey has a mass of 32 kg, and Jonas has a mass of 4

5 kg. When they push off against one another, Jonas acquires a speed of 0.80 m/s .
What is Tracey's speed?
Express your answer to two significant figures and include the appropriate units.
Physics
2 answers:
satela [25.4K]3 years ago
7 0

Mass of Tracey M1 = 32 kg

Mass of Jonas M2 = 45 kg

Initially both were at rest

so V1i = V2i =0

after pushing each other Jonas speed V2f = 0.80 m/s

we need to find out final speed of Tracy

Here we can use momentum conservation as no external force is acting here

M1V1i + M2V2i = M1V1f + M2V2f

32(0) + 45(0) = 32 V1f + 45(0.80)

0 = 32 V1f + 36

-36 = 32 V1f

V1f = - 1.125 m/s

negative sign shows that Tracy will move opposite to the Jonas

so answer in two significant figure would be

V1f = 1.1 m/s

kirill115 [55]3 years ago
6 0
<span>

 = 0.77 m/s hope this helps 

</span>
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MrRissso [65]
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Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and
kow [346]

A) Acceleration: a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first  segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

a_1 = \frac{11-8}{4}=0.75 m/s^2 (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: a_2 = 0

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

a=\frac{0-11}{7}=-1.57 m/s^2

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

u = 0\\a = 0.75 m/s^2\\t=4 s

So the displacement is

s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m

- For the second segment, we have

u = 11 m/s\\a = 0\\t=15 s

So the displacement is

s_2 = (11)(15)+0=165 m

- For the third segment, we have

u = 11\\a = -1.57 m/s^2\\t=7 s

So the displacement is

s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

v=\frac{d}{t}

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

v=\frac{209.5}{26}=8.06 m/s (north)

Learn more about accelerated motion:

brainly.com/question/9527152

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3 years ago
a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

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

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

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Svetach [21]

Answer:

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Explanation:

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Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
3 years ago
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