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dem82 [27]
4 years ago
7

a ball rolling down a hill was displaced 21.9 m while using uniformly accelerating from rest, If the final velocity was 7.14 m/s

, what was the rate of acceleration?
Physics
1 answer:
erastova [34]4 years ago
3 0

Answer:

a = 1.16 m/s²

Explanation:

In order to find the rate of acceleration of the ball, we will use third equation of motion, as follows:

2as = Vf² - Vi²

where,

a = rate of acceleration = ?

s = distance covered by the ball = 21.9 m

Vf = Final Velocity of the ball = 7.14 m/s

Vi = Initial Velocity of the ball = 0 m/s (Since, the ball started from rest)

Therefore,

2(a)(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

<u>a = 1.16 m/s²</u>

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A. An object takes 5 s to reach the
Komok [63]

Answer:

g = 0.4 m/s²

Explanation:

Given the following data;

Height = 5 meters

Time = 5 seconds

To find the acceleration due to gravity (g) on the planet;

Mathematically, the maximum height of an object is given by the formula;

H = ½gt²

Where;

H is the height measured in meters.

g is the acceleration due to gravity.

t is time measured in seconds.

Substituting into the formula, we have;

5 = ½ * g * 5²

5 = 0.5 * g * 25

5 = 12.5 * g

g = 5/12.5

g = 0.4 m/s²

5 0
3 years ago
A child with a weight of 110 N swings on a playground swing attached to 2.00 m long chains. What is the gravitational potential
Ierofanga [76]

Answer

given,

Weight of the child = 110 N

length of the swing,L = 2 m

now, calculating the potential energy when the string is horizontal

  Potential energy = m g h

 now, h = L (1 - cos θ)   where θ is the angle made by the string with the vertical.

     PE = m g L (1 - cos θ)

   when rope is horizontal θ = 90°

     PE = 110 x 2 (1 - cos 90°)

    PE = 220 J

now, calculating potential energy when string made 25° with horizontal

PE = m g L (1 - cos θ)

   when rope is horizontal θ = 25°

     PE = 110 x 2 (1 - cos 25°)

     PE = 20.61 J

5 0
4 years ago
What would be the best way for her to do this?
andrew-mc [135]
Need more info plz :)


5 0
4 years ago
The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhe
WARRIOR [948]

Answer:

6.8 mm

Explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet,m_b=10 g=10\times 10^{-3} kg

1 kg=1000 g

Total mass of system,M=m+m_a=10+10\times 10^{-3}=10.01kg

Speed of bullet,u=300 m/s

\theta=30^{\circ}

By law of conservation of momentum

m_bucos\theta=Mv

v=\frac{m_bvcos\theta}{M}=\frac{0.01\times 300cos30^{\circ}}{10.01}=0.259m/s

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

\frac{1}{2}Mv^2-0=Mgh-0

\frac{1}{2}(10.01)(0.259)^2=10.01\times 9.8 h

Where g=9.8 m/s^2

h=\frac{(0.259)^2}{2\times 9.8}=0.0034m

1m=100 cm

h=0.0034\times 100=0.34 cm

Distance traveled by block=d=\frac{h}{sin\theta}=\frac{0.34}{sin30^{\circ}}=0.68 cm=6.8 mm

1cm=10 mm

4 0
4 years ago
What does the graph to the right represent?
Nastasia [14]
C the thermal equilibrium

4 0
4 years ago
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