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Flura [38]
2 years ago
11

How does the table show that the balloon went downwards ?

Physics
1 answer:
o-na [289]2 years ago
6 0

Answer:

I am going to guess it shows that the balloon is going downwards because the speed of rise is in the negatives for the last 2.

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What type of System interact with its environment ​
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3 years ago
Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe
Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

f=\frac{1}{T}

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

f=\frac{5 cycles}{1 sec}=5 Hz

It follows than the period of the wave is:

T=\frac{1}{f}=\frac{1}{5}s

Therefore, the correct statement is

The period is 1/5 second, and the frequency is 5 Hz.

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3 0
3 years ago
If a Stone with an of original velocity of zero is falling from a ledge and takes eight seconds to hit the ground what is the fi
DaniilM [7]
Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.
3 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
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