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3 years ago

15

While in empty space, an astronaut throws a ball at a velocity of 11 m/s. What will the velocity of the ball be after it has tra

veled 5 meters?

2 answers:

iogann1982 [59]3 years ago

7 0

Answer:

Velocity of the ball will be 11 m/sec after traveling 5 meter

Explanation:

We have given that velocity of ball is 11 m/sec in space

We have to find the speed of the ball after it travels 5 meter

We know that in space there is no force acting

And if there is no force in the ball then velocity of the ball will never change, that is the velocity of the ball is always constant

So the velocity of the ball will be 11 m /sec after traveling 11 m/sec

Mashutka [201]3 years ago

6 0

The velocity will be 11 m/s

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As players speed up, their kinetic energy will?

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Kinetic energy is proportional to the square of the speed. So when anything or anybody speeds up, its kinetic energy increases.

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3 years ago

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A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien

Neporo4naja [7]

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

The mass of the metal rod is $m = 0.12 \ kg$

The current on the rod is $I = 4.1 \ A$

The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

The coefficient of kinetic friction is $\mu_k = 0.18$

The kinetic frictional force is $F_k = 0.212 \ N$

The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as

$F = B * I * L$

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

$F_ k = B* I * L$

=> $B = \frac{F_k}{L * I }$

=> $B = \frac{0.212}{ 6.3 * 4.1 }$

=> $B = 8.20 *10^{-3} \ T$

7 0

3 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

$\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s$

4 0

4 years ago

Mass and energy are alternate aspects of a single entity called mass-energy. The relationship between these two physical quantit

Sever21 [200]

Answer:

Explanation:

ΔE = Δm × c^2

where,

ΔE = change in energy released with respect to change in mass

= 1.554 × 10^3 kJ

= 1.554 × 10^6 J

Δm = change in mass

c = the speed of light.

= 3 × 10^8 m/s

Equation of the reaction:

2H2 + O2 --> 2H2O

Mass change in this process, Δm = 1.554 × 10^6/(3 × 10^8)^2

= 1.727 × 10^-11 kg

The change in mass calculated from Einstein equation is small that its effect on formation of product will be negligible. Hence, law of conservation of mass holds correct for chemical reactions.

8 0

3 years ago

A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density

NISA [10]

Answer:

a) $0.983 \frac{J}{m^3}$

b) $u_E =7.329x10^-3 \frac{J}{m^3}$

Explanation:

The energy density is "the energy per unit volume, in the electric field. The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

$B=\sqrt{2\mu_o u}$

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

$u=\frac{B^2}{2\mu_o}$

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

$B=\frac{\mu_o i}{2\pi R}$

Replacing this on the formula for u we have:

$u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2$

And simplyfing we got:

$u=\frac{\mu_o i^2}{8\pi^2 R^2}$

Replacing the values given we have:

$u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}$

Part b

The density current is given by this formula $J=i/A$ and the resistance by $R=\frac{\rho l}{A}$

If we use the equation for the energy density we have this:

$u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2$

And replacing the values given we have:

$u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}$

4 0

4 years ago