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Aloiza [94]
3 years ago
15

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

hanges, each taking place in a time interval 11.3 s . What are the average acceleration in each interval?
1.) At the beginning of the interval the astronaut is moving toward the right along the x-axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.30 m/s.
a = ___________-??
2.) At the beginning she is moving toward the left at 5.30 m/s. and at the end she is moving toward the left at 15.0 m/s.
a =_________ ??
3.) At the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.
a = _________??
Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer

given,

time interval = 11.3 s

a) initial velocity, vi = 15 m/s

   final velocity, v_f = -5.30 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-5.30 -15}{11.3}

             a = -1.79 m/s²

   the direction is along left side

b)  initial velocity, vi = -5.30 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(-5.30)}{11.3}

             a = -0.858 m/s²

   the direction is along left side

c) initial velocity, vi = 15 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(15)}{11.3}

             a = -2.65 m/s²

   the direction is along left side

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

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

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
2 years ago
What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

4 0
2 years ago
A boy throws a ball up into the air with a speed of 8.2 m/s. The ball has a mass of 0.3 kg. How much gravitational potential ene
diamong [38]
We can use the law of conservation of energy to solve the problem.

The total mechanical energy of the system at any moment of the motion is:
E=U+K = mgh + \frac{1}{2}mv^2
where U is the potential energy and K the kinetic energy.

At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
E_i = K_i =  \frac{1}{2}mv^2 =  \frac{1}{2}(0.3 kg)(8.2 m/s)^2=10.09 J

When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
E_f = U_f
But the mechanical energy must be conserved, Ef=Ei, so we have
U_f = K_i
and so, the potential energy at the top of the flight is
U_f = K_i = 10.09 J
7 0
3 years ago
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The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

F_{new} = 4F_{old}

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

#SPJ4

8 0
1 year ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
2 years ago
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