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EleoNora [17]
4 years ago
12

I have these questions I need to find the acceleration, Please explain how to do them, Don't have to explain all xD

Physics
1 answer:
34kurt4 years ago
7 0
1. a=(v-v0)/t
a=14 m/s / 2s
a=7 m/s^2
2. <span>a=(v-v0)/t
</span>a=(30 m/s - 0 m/s) / 12 s
a=2.5 m/s^2
3. <span>a=(v-v0)/t
a=(37 m/s - 22 m/s) / 2 s
a= 7.5 m/s^2
4. </span><span>a=(v-v0)/t
a=(12 km/s - 0 km/s)/8 s
a=1.5 km/s^2
etc</span>
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A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1
vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

\rho = \frac{m}{V}

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

m_b = 3Kg

r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3

In equilibrium, the entire system is equal to zero, therefore

\sum F = 0

F_g +F_h-F_b = 0

Where,

F_g = Weight of balloon

F_h = Weight of helium gas

F_b = Bouyant force

Then we have,

mg+V\rho g -V\rho_a g = 0

\rho = \rho_0-\frac{m}{V}

Replacing the values we have that

\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}

\rho = 0.978kg/m^3

Now by ideal gas law we have that

PV=nRT

P\frac{\rho}{m} = nRT

P = \rho \frac{n}{m}RT

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

P = \frac{\rho}{M_0} RT

P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K

P = 619995.7Pa

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

5 0
3 years ago
A soccer ball is kicked with a speed of 22m/s at an angle of 35.0º above the horizontal. If the ball lands at the same level fro
Lisa [10]

Answer:

2.57 seconds

Explanation:

The motion of the ball on the two axis is;

x(t) = Vo Cos θt

y(t) = h + Vo sin θt - 1/2gt²

Where; h is the initial height from which the ball was thrown.

Vo is the initial speed of the ball, 22 m/s , θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²

We want to find the time t at which y(t) = h

Therefore;

y(t) = h + Vo sin θt - 1/2gt²

Whose solutions are, t = 0, at the beginning of the motion, and

t = 2 Vo sinθ/g

  = (2 × 22 × sin 35°)/9.81

  = 2.57 seconds

6 0
3 years ago
If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra
igomit [66]

Answer:

\frac{E}{E_c} =3.125

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

E_c=\frac{1}{2} mv^2

The equivalence between mass and energy established by the theory of relativity is given by:

E=mc^2

This formula states that the equivalent energy E can be calculated as the mass m multiplied by the speed of light c squared.

Where c is approximately 3\times 10^{8} m/s

Hence:

E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J

E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125

4 0
3 years ago
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0
4 years ago
why Galileo's thought experiment about the ball rolling forever on a level surface could only be a thought experiment.
erma4kov [3.2K]
Because in reality there are frictional forces acting on the ball, against the direction of its motion. In fact, because of the friction between the ball and the surface, the ball loses little by little its energy, so its velocity decreases and eventually it stops. 
In an ideal world with no friction, there would be no forces acting on the ball, so its energy must be conserved and this means that the ball would continue its motion forever.
8 0
4 years ago
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