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makvit [3.9K]
3 years ago
13

Assume you have measured your little metal sphere's diameter to be 2.500 ± 0.005 cm and its mass to be 63.00 ± 0.05 g.

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

V = (8.2 ± 0.1) cm³ ,    % = 1.2%

Explanation:

The volume of the sphere is

             V = 4/3 π r³

     

The error in volume is

           ΔV = 4 / 3 π r² Δr

           DV = 4 π r² Δr

Let's calculate

         V = 4/3 π (2.500 / 2)³

         V = 8,181 cm³

         ΔV = 4 π 2,500²/4    0.005

         ΔV = 0.098 cm³

         V = (8.2 ± 0.1) cm³

The equation for the percentage error is

     % = ΔV / V 100

      % = 0.1 / 8.2 100

       % = 1.2%

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So the question ask to formulate an expression to represent the amount of air required to inflate the balloon from a radius of R+4 and the expression would be V=16pi/3*(3r^2+12r+16). I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarifications
4 0
3 years ago
8. What is the momentum of a 120-kg professional fullback running<br> across the line at 11.2 m/s?
Alchen [17]

Answer:

134r kgm^-1 or 1344 kg /m

Explanation:

Momentum is is given by:

p=mv

Where:

p is momentum, m is mass in kg and v is velocity in ms−1

p=120kg×11.2 m/ s= 1344 kgms=1344kgm^−1

5 0
3 years ago
A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelera
user100 [1]

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

<u>Vf₁ = 25 m/s</u>

<u></u>

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

<u>s₂ = 90 m</u>

<u></u>

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

5 0
3 years ago
(100 POINTS!!) Short answer about how acceleration is dependent on the speed and direction of an object!!!! please help!!!!!
zhuklara [117]

Answer:

The acceleration of the object is dependent upon this velocity change and is in the same direction as this velocity change. The acceleration of the object is in the same direction as the velocity change vector; the acceleration is directed towards point C as well - the center of the circle.

i hope that helped you!!

Explanation:

8 0
3 years ago
Read 2 more answers
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is p
madreJ [45]

Answer:

Part a)

L = 0.68 m

Part b)

L = 0.63 m

Explanation:

Part a)

As we know that there is no friction in the path

So here we can use energy conservation to find the distance moved by the mass

Initial spring energy = final gravitational potential energy

so we will have

\frac{1}{2}kx^2 = mgL sin\theta

\frac{1}{2}70(0.5)^2 = 2(9.81)(L) sin41

8.75 = 12.87 L

L = 0.68 m

Part b)

Now if spring is connected to the block then again we can use energy conservation

so we will have

\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2

so we will have

\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2

8.75 = 6.43 + 12.87 x' + 35 x'^2

x' = 0.13 m

so total distance moved upwards is

L = 0.5 + 0.13 = 0.63 m

8 0
3 years ago
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