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Harlamova29_29 [7]
2 years ago
12

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

Gravitation.
Physics
1 answer:
yKpoI14uk [10]2 years ago
3 0

Answer:

The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

v=\dfrac{dx}{dt}....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

Put the value of p

F=\dfrac{d(mv)}{dt}

F=m\dfrac{dv}{dt}

Put the value from equation (II)

F=ma

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

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7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in
viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

6 0
2 years ago
A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c
Nana76 [90]

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

F=\mu mg = m \frac{v^2}{r} (1)

where

\mu is the coefficient of friction

m is the mass of the car

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

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

\mu=0.50\\v = 20 mph = 8.9 m/s

So the radius of the curve is

r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when \mu=1.0. We find:

v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph

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Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

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